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Topics - rabinoz

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1
Flat Earth Investigations / More on "13 Miles: 60 ft NOT Hidden".
« on: November 06, 2018, 07:24:05 AM »
This was posted in Flat Earth Media but that is hardly to forum for further discussion so I have started this new topic.

Even with very strong atmospheric refraction, I don't think this should be possible on a globe:



I look forward to seeing this done again across greater expanses. I need to see if this is repeatedly under standard conditions. In fact, I want to do it myself. I have no answer for this and concede this strongly supports a flat earth...for the time being.

I can see globies are ignoring such a great visual scientific experiment. If the experiment does not prove a curve they want nothing to do with it.

I wonder what their argument will be?
Now I'm no meteorologist but an explanation might be atmospheric ducting due to a temperature inversion. This is not uncommon in that region.
The following references might be useful:
Quote
Atmospheric duct

Fata Morgana of Farallon Islands with clearly seen duct


In telecommunications, an atmospheric duct is a horizontal layer in the lower atmosphere in which the vertical refractive index gradients are such that radio signals (and light rays) are guided or ducted, tend to follow the curvature of the Earth, and experience less attenuation in the ducts than they would if the ducts were not present. The duct acts as an atmospheric dielectric waveguide and limits the spread of the wavefront to only the horizontal dimension.

Atmospheric ducting is a mode of propagation of electromagnetic radiation, usually in the lower layers of Earth’s atmosphere, where the waves are bent by atmospheric refraction. In over-the-horizon radar, ducting causes part of the radiated and target-reflection energy of a radar system to be guided over distances far greater than the normal radar range. It also causes long distance propagation of radio signals in bands that would normally be limited to line of sight.

Normally radio "ground waves" propagate along the surface as creeping waves. That is, they are only diffracted around the curvature of the earth. This is one reason that early long distance radio communication used long wavelengths. The best known exception is that HF (3–30 MHz.) waves are reflected by the ionosphere.

The reduced refractive index due to lower densities at the higher altitudes in the Earth's atmosphere bends the signals back toward the Earth. Signals in a higher refractive index layer, i.e., duct, tend to remain in that layer because of the reflection and refraction encountered at the boundary with a lower refractive index material. In some weather conditions, such as inversion layers, density changes so rapidly that waves are guided around the curvature of the earth at constant altitude.
These are also relevant:
         Calculating Ray Bending This gives a simplistic calculation of the lapse rate  needed to cause ducting.
         Ducts More specific discussion of ducts,  with diagrams.
         Marine layer Discusses the "marine layer", common in the Monterey Bay area.
         Monterey Bay National Marine Sanctuary, Regional pressure and temperature effects.
Maybe someone can make something of that material.

2
Most discussion about circumnavigation seems to have been about east to west circumnavigation.
But what about circumnavigation from north to south to the South Pole, north to the North Pole then south to the point of departure.

Circumnavigation by air via both the South Pole and the North Pole has been done a few times. Here are a couple.
Quote from: Guiness Book of Records
First Circumnavigation via both Poles by Aircraft
Captain Elgen M. Long achieved the first circum-polar flight in a twin-engined Piper PA-31 Navajo from 5 November to 3 December 1971. He covered 62,597 km (38,896 miles) in 215 flying hours.
from: First Circumnavigation via both Poles by Aircraft, Guiness Book of Records

Quote from: NY Times
Charles Burton 59 a Pole-to-Pole Explorer
Charles Burton, a British explorer who took part in the first expedition to circumnavigate the globe from pole to pole, died on Monday at his family home in the English village of Framfield in Sussex. He was 59 and had suffered a heart attack, said his brother, Richard.
from: Charles Burton 59 a Pole-to-Pole Explorer, NY Times

See a bit more in, Traveling Directly South? « Reply #1 on: February 07, 2017, 12:48:59 PM »

And you can do it yourself:
Quote
THE TRIP OF A LIFETIME DEPARTS JFK AT 11:00 am, October 26, 2018
If you are the kind who would like to tell your children’s children that you were one of an exclusive group of people who have circumnavigated the Earth over both poles, step aboard our private Airbus A340-300 on October 26, 2018. We call this historic one-time flight “THE POLAR EXPLORER.” Jose Roberto Àldana, Travel Curator

The flight starts in New York. From there, in one giant nonstop leap (6,382 miles - 11 hours, 35 minutes) we fly southwards to Río Gallegos, at the southernmost tip of Argentina. Then across the entire continent of Antarctica, for a spectacular daytime crossing directly over the geographical South Pole (6,663 miles - 12 hours, 20 minutes) to Perth, in Western Australia. Our flight then turns north, for a journey to Beijing in China (4,962 miles - 9 hours, 10 minutes). It then crosses continental Russia and the Siberian Plateau over the geographic North Pole to New York (6,710 miles, 12 hours, 45 minutes), landing back at JFK at around 2:00 p.m., approximately 50 hours later.

See more at: THE TRIP OF A LIFETIME DEPARTS JFK AT 11:00 am, October 26, 2018
And here is the route they'll be flying:
A video project by GreaterSapien documenting my trip on a world record flight circumnavigating both the North and South Poles
https://www.kickstarter.com/projects/1282946600/pole-to-pole-circumnavigation-of-the-globe
The flight from New York has four stops: Rio Gallegos, Argentina, Perth, Australia, Beijing, and back to New York. The info can be found here  (https://www.overthepoles2018.com).
And who knows, if we raise enough, maybe I can entice a big Flat-Earther to come with me?



Regards,
Dens
A YouTube poster, GreaterSapien, has crowdfunded his seat on the flight.
Kickstarter has been 100% funded!
Are any keen flat-earthers going to accompany him?

3
Flat Earth Theory / Gleason's New Standard Map of the World
« on: January 24, 2017, 09:54:56 PM »
For the information of Flat Earthers.

I had been under the impression that "Gleason's map is the New Standard Map of the World" was simply a version of the
"North Polar Equidistant Azimuthal Projection of the Globe".
It is certainly based on it, but was specifically produced by Alex. Gleason as his Flat Earth map.

I have a copy and the text on it reads:
GLEASON’S
NEW STANDARD MAP OF THE WORLD
ON THE PROJECTION OF J. S. CHRISTOPHER,
MODERN COLLEGE, BLACKHEATH, ENGLAND
SCIENTIFICALLY AND PRACTICALLY CORRECT
AS “IT IS”

You will note that it says "ON THE PROJECTION OF J. S. CHRISTOPHER" and is certainly a North Polar Equidistant Azimuthal Projection of the Globe.

Others may have know, but I have recently only found out that Alex. Gleason was a civil engineer who most certainly believed in a Flat Earth and
produced what we now call "Gleason's Map" is his version of the AEP map (as in Rowbotham's ENaG).

Alex. Gleason wrote a couple of books, "Is the Bible from Heaven" and "Is the Earth a Globe" that can be downloaded as one PDF scanned document from:

And this is a link to a 4773 x 6794 pixel version of the map:

But, I have no more information on "J. S. Christopher, Modern College, Blackheath, England" or of "Morden College".
Nevertheless, whatever his credentials, J. S. Christopher seems to have produced a quite accurate map.


1892-new-standard-map-of-the-world
The resolution of this copy is limited by Photobucket.

As far as I am concerned, all this does is to validate Gleason's Map as a Flat Earth map
                      - that was it's intention and it is really just a far more accurate version of the "Ice Ring" map that most people use.


4
Suggestions & Concerns / Why am I falsely accused of spreading lies?
« on: January 13, 2017, 01:19:51 AM »
In a post related to "Conspiracy" I included
Ski is pretty high in "The Flat Earth Society".

SexWarrior replied with this:

<< above quote >>
Please refrain from spreading lies. Ski is a member of another organisation, not this one. Thanks!
Now I do not like being falsely accused of lying (nor of "spreading lies")

I did not say "this one", I said "The Flat Earth Society" and Ski is a member of "The Flat Earth Society".

So, SexWarrior Stop making false accusations!

Look at how long each organisation has been in existence:

The Flat Earth Society (Flat Earth Society) The Flat Earth Society
Early members:
Daniel                    2005-02-14
John Davis            2006-12-06
Lord Wilmore     2006-04-16
Raa                       2006-10-07
Colonel Gaydafi 2007-02-26

So apparently "The Flat Earth Society (Flat Earth Society)" has been in operation for almost 12 years.

But,
The Flat Earth Society (TFES)
Early members:
Parsifal        2012-03-13
SexWarrior 2012-03-15
Snupes       2012-03-16
jroa             2013-08-30

so apparently "The Flat Earth Society (TFES)" has been in operation for less than 5 years.

Now, I am making no claims as to which is the better society, this on is "better" moderated, but has virtually no activity.
The other tends to be too much "Rafferty's rules" at times - but that has nothing to do with the case.

There seems no doubt that "The Flat Earth Society (theflatearthsociety.org)" has been in operation for much longer than "The Flat Earth Society (tfes.org)".
Hence would certainly seem to have the prior claim to the name "The Flat Earth Society".

All I claim is that I was perfectly correct in stating that "Ski is pretty high in 'The Flat Earth Society'", though on reflection I could have said, "in 'The other Flat Earth Society'".

But to be accused of "spreading lies" for that is going too far.
If you two Flat Earth Societies can't get on, nor even decide on unique names, don't take it out on me!
Stop being childish and sort it all out.

<< left out half topic! >>

5
I have posted essentially the same material numerous times here and on The Flat Earth Society, Forum and I have never had a satisfactory answer.

This is what "the Wiki" says (bits about solar eclipse etc, removed for brevity):
Quote from: The Wiki
The Lunar Eclipse
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A Lunar Eclipse occurs about twice a year when a satellite of the sun passes between the sun and moon.

This satellite is called the Shadow Object. Its orbital plane is tilted at an angle of about 5°10' to the sun's orbital plane[1], making eclipses possible only when the three bodies (Sun, Object, and Moon) are aligned and when the moon is crossing the sun's orbital plane (at a point called the node).    . . . . . . .  A lunar eclipse can be seen from the entire half of the earth beneath the moon at that time.
   . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The shadow object is never seen because it orbits close to the sun.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
It is estimated that the Shadow Object is around five to ten miles in diameter. Since it is somewhat close to the sun the manifestation of its penumbra upon the moon appears as a magnified projection. This is similar to how during a shadow puppet show your hand's shadow can make a large magnified projection upon your bedroom wall as you move it closer to the flashlight.
From The Lunar Eclipse

This is my interpretation of that geometry. In this diagram the size of the objects has been enlarged (or they would be almost invisible), but the locations are approximately to scale:


With the "shadow object" so small, it is quite impossible for the it to cast any significant shadow on the moon. Almost all of the sunlight will shine around it.

If my interpretation of the geometry or light paths is incorrect, I would love to be informed, but please no massive refraction or magnification in the atmoplane, there is no atmoplane 5,000 km up!

So, I claim that "the Wiki" explanation of the Lunar Eclipse is completely incorrect, so what is the true cause of a Lunar Eclipse.

Some will I ask why I am asking the same question over and over. The answer to that is simple - it has not yet been answered.


[1] If the "shadow object" can never be seen, how was the inclination of its orbit determined, for we are told "Its orbital plane is tilted at an angle of about 5°10' to the sun's orbital plane"?
      I can guess, that's simply been "borrowed" from the measured orbital inclination of the moon by astronomers!

6
"TFES Wiki" contains the following claims:
Quote
Sinking Ship Effect
It is proven that the ship does not sink behind a hill of water, but that it is actually perspective which hides it. This demonstrates that the earth is not a globe. There have been experiments where half-sunken ships have been restored by simply looking at them through telescopes, showing that they are not actually hiding behind "hills of water".

This is presented in two parts:
  • The claim " that the ship does not sink behind a hill of water"
    The first claim is "It is proven that the ship does not sink behind a hill of water, but that it is actually perspective which hides it."

    I can find claims to this effect, but no proof, simply the statement
              "that it is actually perspective which hides it.", followed by Rowbotham''s explanation on his "proof",
               which pre-supposes that the earth is flat. In other words, all we have is "If the earth were flat, this might explain the 'sinking ship' phenomenon."
    Then the statement "This demonstrates that the earth is not a globe."
    I do not regard any of this as proof, as it pre-supposes that the earth is flat.

  • Then the claim "half-sunken ships have been restored"
    Then simply a bald statement "There have been experiments where half-sunken ships have been restored by simply looking at them through telescopes".
    The statement "There have been experiments" is exactly the thing that globe supporters are criticised for! Where are always told to show "the evidence"!

So, please publish the evidence that "half-sunken ships have been restored by simply looking at them through telescopes"

And boats that are simply too small to see on the horizon "restored" by a long zoom lens will not be considered as evidence.


7
Flat Earth Theory / Why can't the Sun be seen at midnight?
« on: October 30, 2016, 05:24:42 AM »
"The Wiki" explains "day and night" cycles in the well-known fashion as in
Quote
How do you explain day/night cycles and seasons?
Day and night cycles are easily explained on a flat earth. The sun moves in circles around the North Pole. When it is over your head, it's day. When it's not, it's night. The sun acts like a spotlight and shines downward as it moves. The picture below illustrates how the sun moves and also how seasons work on a flat earth. The apparent effect of the sun rising and setting is usually explained as a perspective effect.


An animation of the day/night cycle in FET

Now just as an example take the sun position as shown on the "The sun's orbit at equinox" circle, above the equator at long 180°. For a position on earth along longitude 0° to have darkness at midnight the sun 180° away must not be vidible, presumably due to the "finite transparancy of the air".

But in Rowbotham's explanation of the "Retrograde motion of the Planets" on pp 322. 323 we have this explanation:
Quote from: Parallax
STATIONS AND RETROGRADATION OF PLANETS

The planets are sometimes seen to move from east to west, sometimes from west to east, and sometimes to appear stationary, and it is contended that “the hypothesis of
the earth’s motion is the natural and easy explanation; and that it would be in vain to seek it from any other system.”

To those who have adopted the Newtonian theory the above language is quite natural; but when the very foundation of that system is proved to be erroneous, we must seek for the cause as it really exists in the heavens, regardless of every hypothesis and consequence. Careful observation has shown that the advance, apparent rest, and retrogradation of a planet is a simple mechanical result.

All the orbits are above the earth; and whenever a spectator stands in such a position that a planet is moving from right to left, he has only to wait until it reaches the end or part of its orbit nearest to him, when, as it turns to traverse the other side of the orbit, it will, for a time, pass in a direction to which the line of sight is a tangent.

A good illustration will be found in an elliptical or circular race-course. A person standing at some distance outside the course would see the horses come in from the right, and pass before him to the left; but on arriving at the extreme arc they would for a time pass in the direction of, or parallel to, his line of sight, and would, therefore, appear for a time not to progress, but on entering the other side of the course would appear to the spectator to move from. left to right, or in a contrary direction to that in which they first passed before him. The following diagram, fig. 99, will illustrate this.



FIG. 99.

Let S be the place of the spectator. It is evident that a body passing from A to P, would pass him from right to left; but during its passage from P to T it would seem not to move across the field of view. On arriving, however, at T, and passing on to B, it would be seen moving from left to right; but from B to A it would again appear to be almost stationary.

From ZETETIC ASTRONOMY, ZETETIC AND THEORETIC DEFINED AND COMPARED.

Now clearly, Rowbotham is saying that we can see the planets when they are at all positions around the 360° of their orbit, otherwise they could never show retrograde motion. So I have to wonder, "If we can see the planets when they are on the other "side" of the earth, why can't we see the sun, which is almost infinitely brighter than any planet?"

This explanation of "retrograde motion" differs markedly from the explanation in "the Wiki", so we have to ask
if Rowbotham is so wrong in this, what other parts of Rowbotham's writings are also simply incorrect.

8
Flat Earth Community / How does TFES explain.
« on: October 26, 2016, 12:30:19 AM »
On a recent thread I posted that:
. . . . . . . .
(The Flat Earth had) no reasonable explanations for so many observations:
          The sharp horizon, especially at sea on a clear day,
          Lunar eclipses,
          Sunrise and sunsets,
          The constant size of the sun and almost constant of the moon from rising to setting,
          The directions and times of sunrises and sunsets,
etc, etc. 
And, before you ask, I have looked up "the Wiki" and read quite a lot ENAG.

The "highly informative" reply was:
. . . . . . . . . .
All of the observations are accounted for in FET. You say that you have read some of the reference material. I would suggest going back and doing it again.
. . . . . . . . . .
I have found very little in the Q&A and to me, at least, "the Wiki" explanations are either quite impossible or completely unconvincing.

Does anyone have any better suggestions?

I know that there are really 5 questions here, so I don't expect anyone to try to answer them all.

Some individual topics could then possibly go into "Flat Earth Debate".

9
There has been similar material in another thread, but it never settled the crucial question!

Junker posted this in a reply to another post:
Quote
Ships over the horizon reappear when you look at them through a telescope.
This one is a legit argument and those who have completed the experiment have confirmed it.

Where is this experiment that confirms this?

It MUST show that Ships over the horizon reappear when you look at them through a telescope.

The critical point here is disappearing OVER the horizon, not disappearing simply because they are small boats that are too small to see.

In other words a ship large enough to clearly see and resolve into parts must disappear OVER the horizon as we know they do, and has been know for millenia.

10
Flat Earth Theory / How does the moon show the same face everywhere?
« on: June 19, 2016, 09:17:18 PM »
We know that the moon shows almost exactly the same face to all observers, no matter where the are on earth or the position of the moon in the sky.

"the Wiki" describes the moon as
Quote from: the Wiki
Moon
The moon is a rotating sphere. It has a diameter of 32 miles and is located approximately 3000 miles above the surface of the earth.

Some viewers see the moon directly overhead, others see it on the horizon.
These viewers would seem to be viewing completely different faces, yet we know that everyone sees the SAME face.

How is this possible?

11
The last post on this whole site seems to be:
I'm for reunification in general, but I don't see it happening. Our setup is a well-oiled machine. It would be nice to get more noobs, but it isn't worth sacrificing how well this place is run to accomplish that. We just have better stuff.

It is true that this site is "a well-oiled machine" and better moderated.

It might also be true that "We just have better stuff", but somehow is seems to have gone to sleep!

You know full well that I am not sympathetic to the idea of a "Flat Earth", nevertheless I believe sites like this perform a very important function.

The rubbish on YouTube shows that most "Flat Earthers" there are simply people not smart enough to understand the Globe and know little of the intricabcies of the Flat Earth model.

Sites like this are necessary as somewhere people can go to get a more reasoned view, and see how the Flat earth model is supposed to work.

Mind you I don't think it does, but that's another story for another place.

I'll climb down off my soapbox and see what happens.

12
There have been a number of threads on Geodetic Surveyors and how they "prove the earth is a Globe". So many "shot them down" saying that Geodetic Surveyors did not "prove curvature", possibly not, but as I have stated so often Geodetic Surveyors measured the earth and have done so with increasing accuracy for centuries. The sizes of all countries and the lat/long coordinates of all features have been measured and can readily be ascertained. The most compact source of this information is in maps (some very accurate, others less so) and atlases (often not very accurate, but usually quite close enough).

And it is these measurements that show us the "shape of the earth".

The following assumes that at a given latitude, a degree of longitude is the same all around the globe.

This seems quite consistent with all evidence[1] and with what is in "the Wiki" on Finding your Latitude and Longitude.

The following table gives the width of one degree (under the heading "km/deg") at various latitudes in both the northern and southern hemispheres, the circumference of the earth (the distance for 360°) from the map reading at each latitude, the circumference at that latitude based on a spherical earth (yes, I know it's not a perfect sphere!) and the circumference at that latitude based on a Flat Earth Ice Wall Map.

The "Flat Earth Circumference @ Latitude" is based on the 24,900 mile overall diameter of the "known earth" as in the Wiki, The Ice Wall. The circumferences are then simple "pro-rated" as the meridians on the "FE Ice Wall map" are simply radial lines.

Latitude
   

km/deg @ Lat
   
Map
Circum@Lat
   
Ideal Globe
Circum@lat
   
Flat Earth
Circum@lat
   
Source of "map data"
51.0°
   
70.3 km/deg
   
25,300 km
   
25,200 km
   
27,400 km
   US 1887 map
43.0°
   
81.7 km/deg
   
29,400 km
   
29,300 km
   
33,000 km
   US 1887 map
35.0°
   
91.4 km/deg
   
32,900 km
   
32,800 km
   
38,600 km
   US 1887 map
0.0°
   
109.7 km/deg
   
39,500 km
   
40,100 km
   
63,200 km
   Times Atlas map
-20.0°
   
102.1 km/deg
   
36,700 km
   
37,700 km
   
77,200 km
   Times Atlas map
-34.0°
   
92.0 km/deg
   
33,200 km
   
33,200 km
   
87,100 km
   1855 Australian map
-45.0°
   
79.2 km/deg
   
28,300 km
   
28,300 km
   
94,800 km
   Times Atlas map
-55.0°
   
65.5 km/deg
   
23,600 km
   
23,000 km
   
101,800 km
   Times Atlas map
The 1887 US survey map and the 1855 Australian map are very high resolution accurate maps, but the Times Atlas is not such a large scale and not as accurate. Also the figures are scaled (quite accurately) from scanned paper maps, so very high accuracy is not expected. Nevertheless most of the circumferences are within 1% of the expected value for the globe (The "Times Atlas" is a bit out at high southern latitudes - not unexpected for a flat map).

These measurements can be repeated anywhere on earth you like. In many cases, if you have a straight E-W stretch of road you can verify the results yourself. No great accuracy is needed, as the differences between a globe and the flat earth are massive!

But, unless you seriously doubt maps that have been in use for many years, the circumference of the earth gets less as we move North or South from the equator.
To me (as William Carpenter might have said): "this is an incontrovertible proof that the Earth is a globe."


[1] Any navigator would think it ridiculous to suggest otherwise.

13
Flat Earth Theory / How does a Full Moon appear Full for everyone?
« on: April 18, 2016, 12:22:27 PM »
Before I start, when I say "for everyone", I do mean for all those over the half of the earth that can see the moon.

We are told
Quote from: the Wiki
The Phases of the Moon
When one observes the phases of the moon he sees the moon's day and night, a shadow from the sun illuminating half of the spherical moon at any one time.
The lunar phases vary cyclically according to the changing geometry of the Moon and Sun, which are constantly wobbling up and down and exchange altitudes as they rotate around the North Pole.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
When the moon is above the altitude of the sun the moon is fully lit and a Full Moon occurs.
In the diagram below I have drawn how I picture the geometry at the time of a full moon (some diagrams in the Wiki might help) of the earth, sun and moon. I have not drawn the moon above the sun, as the time of a full moon the moon would be around 20,000 km from the sun, so a few tens of kilometres could hardly make a difference! If the moon were much higher it would appear much smaller at the time of a full moon - and it certainly does not.
In this diagram horizontal and vertical distances are to scale, but the object (and people) sizes are exaggerated, or else they would be quite invisible.

OK, so you have the half the moon illuminated by the light from the sun. But, it is illuminated on the side!. The observer directly underneath is looking straight up and clearly sees only half the side facing him illuminated, that is sees only a HALF MOON, not a Full Moon.

The other observer, for which the sun would be just setting and the moon rising (or vice versa), sees most of the part of the moon facing him as illuminated, so sees a nearly full moon.

But, we know for a fact that the phase of moon does not change (substantially) throughout the night or for observers in different locations, and not as appears here
almost a full moon for those where the moon is near the horizon and only a half moon for those directly under the moon.

Please explain where I am mistaken, because this is how I interpret what is said in the Wiki.

14
The Wiki gives the sun height as the same as the distance from the equator to latitude 45° North at "approximately 3,000 miles" in

Sun's Distance - Zetetic Cosmogony
Thomas Winship, author of Zetetic Cosmogony, provides a calculation demonstrating that the sun can be computed to be relatively close to the earth's surface if one assumes that the earth is flat --
Quote
On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.
This is illustrated in this diagram from Modern Mechanics - Oct, 1931:

Voliva's Flat Earth Sun Distance.
This is also shown in the Wiki under Distance to the Sun under the section Sun's Distance - Modern Mechanics.

But this calculates the height from only ONE location, Latitude 45°. In would seem that we would get a more accurate result by taking measurements from a number of different latitudes and averaging the result.

In Finding your Latitude and Longitude the Wiki says:
Quote
Latitude
To locate your latitude on the Flat Earth, it's important to know the following fact: The degrees of the Earth's Latitude are based upon the angle of the sun in the sky at noon equinox.
That's why 0˚ N/S sits on the equator where the sun is directly overhead, and why 90˚ N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45˚ in the sky. The angle of the sun past zenith is our latitude.
Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.

So this quote from the Wiki tells how far we would be from the equator and what the sun's angle would be at any latitude.

This enables us calculate the sun's height from
h = d x tan(Θ), where "d" is the distance from the equator, d = 69.5 x Θ,
and the sun's elevation Θ = (90° - Latitude),  as illustrated in:
The sun's height, calculated using this method for a number of latitudes is shown in the following table:
Latitude 
Sun Elevation, Θ   
Distance from Equator, d   
Sun Height, h   
7.2° N   
82.8°
500 miles
3,961 miles
15° N   
75°
1,043 miles
3,891 miles
45° N   
45°
3,128 miles
3,128 miles
75° N   
15°
5,213 miles
1,397 miles
85° N   
5,908 miles
517 miles
Here we see that at a latitude of 45° N (3,128 miles from the equator) the sun's height comes out to be 3,128 miles, more or less as expected.
But, at all other latitudes we get quite different results ranging from 3,961 miles at 7.2° N from the equator to only 517 miles at 85° N.

All of the figures used have come from the Wiki, and the calculation is based on the method from the Wiki.
So can someone explain why these calculations (using the method from the Wiki) give such different figures for the sun's height?
I know the same point has been raise numerous times, but has never been given a reasonable answer.

15
The Geodetic Surveyor Jesse Kozlowski made this video describing just what a Geodetic Surveyor does and how in the end these surveyors have effectively "measured the earth" and shown that it simply cannot be a flat plane. This might seem to contradict the "Hundred Proofs"
Quote from: the Wiki
A hundred proofs the Earth is not a globe
Surveyors' operations in the construction of railroads, tunnels, or canals are conducted without the slightest "allowance" being made for "curvature," although it is taught that this so-called allowance is absolutely necessary! This is a cutting proof that Earth is not a globe.
So it might be worth finding out what a "Geodetic Surveyor" does that is different from the job a "Plane Surveyor: does.
The video is aimed at those looking into the possibility of the Flat Earth, and does dissect some FE videos that have been on Youtube, showing where they are quite mistaken.

His presentation is quite reasoned and quiet, with none of the histrionics we find on many Youtube FE videos.
The results he presents go right back to the earliest days of Geodetic Surveying, way before there was any thought of the current FE movement and conspiracy ideas.

But, I have to warn any viewers that it quite long, though nothing like the marathon sessions needed for some of "The biggest lie" videos.


I found it quite worth watching, though I could be classed as seriously biased!

16
My contention is that the accepted dimensions of the earth simply do not fit of a flat surface.

In the post https://forum.tfes.org/index.php?topic=4557.msg88728#msg88728 I tried to argue the following:

The accepted distance from the equator to the north pole is 10,000 km, (give or take a few km).
The accepted circumference at the equator (flat or globe) is 40,000 km, (give or take a few km).

On any flat earth map I have seen the radius of the equator circle is simply the distance from the equator to the north pole.

But on any round disk a circle (the equator circle) the circumference is 2 x π x radius (of the equator circle) or 62,832 km.

So, by my reckoning on the real earth the distance around the equator is 40,000 km,
but if the earth were flat, to fit with the 10,000 km
equator to pole distance distance around the equator would have to be 62,832 km.

This seems to imply that the earth with these measurements simply will not fit on a plane surface.

There are a number of ways to justify these figures (or very close ones), some from flat earth sources, but I will omit these for brevity here.
You can look up my previous posts for some of these.

Basically the only response I have had is one from Tom Bishop that simply said "evidence".  In the original post, and in the answer to Tom Bishop I gave some more evidence. If someone has some other dimensions, we can discuss those, but I do believe I have enough support even some in Zetetic Astronomy, Earth Not A Globe and other Flat Earth sources.
Now, unless someone comes up with some dimensions that differ markedly from mine,
the only conclusion that can be drawn is that THE EARTH is NOT FLAT.
I am quite prepared to accept that I have presented this in a boring fashion, or badly,
but I do firmly believe that there is a real case to answer!

[Edit - a little better wording]

17
This attempts to show that the accepted dimensions of the Earth simply will not fit on a flat surface.

It seems fairly well accepted(1) that the
distance from the equator to the north pole is (close to) 10,000 km and that the

equatorial circumference of the earth is (close to) 40,000 km(2).
So that the
(equatorial circumference) = 4 x (distance from the equator to the north pole).

But, on any round flat earth map(3) I have seen the distance from the "equator to the north pole" is the radius of the equatorial circle and we know that the
(circumference of equatorial circle) = 2 x π x (distance from the equator to the north pole) or 62,832 km

There is a big discrepancy here.  Put simply:
The accepted dimensions of the earth do not fit on a flat surface.
From what I can see the only "practical" object that these dimension can fit is a sphere! (4)


(1)One way to derive these dimensions is the original definitions of the Nautical Mile as one minute of either longitude or latitude.
So the equator to pole distance becomes 90° x 60' x 1Nm/min = 5,400 Nm or 10,000.8 km
and the circumference around the equator becomes 360° x 60' x 1Nm/min = 21,600 Nm or 40,003.2 km

(2)I know that the accepted figures are 40,075 km and 10,002 km resp, but I am using rounded numbers for simplicity.

(3)The Bi-Polar model has been suggested elsewhere, but I really do not see how that can help.

(4)Redefining π as 2 might help - bit that is a bit way out for me!.

18
Whatever the shape of the earth, we all live in the same one,

So from "Down Under", a Merry Christmas to you all!

And to the many who might celebrate differently, happy Holiday and Peace of Earth!

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