FE radius (UAFE estimate)
« on: August 11, 2023, 07:04:34 PM »
https://wiki.tfes.org/Eratosthenes_on_Diameter

Therefore we can take the distance of 500 miles, multiply by 25, and find that the radius of the flat earth is about 12,250 miles. Doubling that figure for the diameter we get a figure of 25,000 miles.


12,250 mi = 19,600 km

This can't be right, not even close.

The lower bound would be R = 6,363.63 km, the upper bound (the RE distance from the equator to the NP) is 10,000 km.

More details here:

http://www.ilya.it/chrono/pages/erdmessungen.htm

Offline jimster

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Re: FE radius (UAFE estimate)
« Reply #1 on: August 16, 2023, 08:11:08 PM »
The FE map relates to the RE globe mathematically as a projection. If you take the surface of a sphere and project it onto a 2d disk, the only distances that are preserved, i.e. both the same, are distances along the north/south radians. Doing this gives a distance from north pole to equator of 6,215 mi. This matches gps, celestial navigation, time/speed/distance, google maps distance tool, etc etc etc. I have never seen a FE map with a scale, but this is one way to do it. If you use that scale on the FE map, the east/west distances are too long (vs all conventional sources) in the northern hemisphere and too short in the southern, which matches the distortion expected when projecting a sphere onto a disk. This is why on the FE map, the US is narrower than Australia, while in actuality, Australia is about 300 mi less east to west distance.

If you postulate that the edge of FE is the same distance from equator as from north pole to equator, this gives FE radius as 4 times 6215, or 24,860 mi (approximately). As I said, this matches RE geometry and observation, at least along direct north/south radians. But ... problems.

The FE radius is unknown. First, because FE maps don't have scales, so we don't know if they intend to match RE distance directly north/south. Indeed, one prominent FE poster said that no one knows the distances across oceans because you can't crawl across them with a ruler and that is the only way to measure. Second, because FE does not know where the southern edge of FE is. Some say it is an infinite plane, some say there are lands beyond , and no one knows how the dome meets the surface or where exactly it is down there. Until it is ascertained where the edge is and what the scale is, how can we know the radius?

If you take the FE north/south distances as the same as the RE distances, and you take the edge as the circle formed by the top down mathematical projection of sphere onto disk, then the diameter of FE is equal to the circumference of RE. But that requires knowing that the scale of the north/south lines on the projection is the same as RE and that the edge is where the circle would be with a mathematical projection of RE onto FE. Can we say that FE agrees these distances are the same, there is an edge, and it is the same distance from the equator as the north pole?

Seems to me that either the FE north/south distances are the same as RE, in which case Australia is too wide by hundreds of mile, or if the north/south scale is not the same. perhaps unknown, and the edge is not the same distance from equator as from north pole to equator, then how can we ever calculate the radius of FE?

If you take the FE north/south scale as equivalent to RE, then the FE east/west distances do match observations and RE does match. If you don't know the scale of the map, you don't know the distance across ocean, and you don't know where the edge is, how can you ever know the diameter of FE? I submit that either FE does not match observations, or we are all so ignorant that we may never know.

Or you could do the math for Eratosthenes experiment while assuming FE and get numbers that don't match anything, I guess. Then you have to explain everything from gps to airline schedules etc etc etc not matching. That will take a huge conspiracy involving geographers, cartographers, astronomers, navigators, geodetic surveyors, etc etc etc, all of which must either be sinister of unable to do math. I have heard FEs take the position that scientists are stupid and/or evil and actually no one knows because FE is a young science (not true) and does not have the resources to figure it out (true).

The last possibility is that our powers of observation are all distorted in the exact way to make FE look like RE, perhaps by some uber Einstein n-dimensional math. That leaves you with RE for all practical purposes and FE as a mathematical concept of ultimate truth. Not very useful or satisfying, except to be prideful or perhaps to make your preexisting beliefs (religion) plausible.

To know the diameter of FE, you have to have a map with a scale and know where the edge is. So far, I have only heard speculation on this from FE, but if you can define it, you can calculate. Then you can explain why it does not match observations, but that may be a different thread.

I welcome anyone pointing out errors of math or logic, and especially welcome anyone with a coherent logical explanation of FE diameter that matches observations.
I am really curious about so many FE things, like how at sunset in Denver, people in St Louis see the dome as dark with stars, while people in Salt Lake City see the same dome as light blue. FE scientists don't know or won't tell me.

Re: FE radius (UAFE estimate)
« Reply #2 on: August 17, 2023, 05:23:51 AM »
Whoever wrote the works attributed to Erathosthenes used spherical triangles:

https://web.archive.org/web/20080214044528im_/http://geocities.com/levelwater/shadow01.gif
https://web.archive.org/web/20080214044528im_/http://geocities.com/levelwater/eratostenesegments.gif
https://web.archive.org/web/20080214044528im_/http://geocities.com/levelwater/tangentedecurva02.gif
https://web.archive.org/web/20080214044528im_/http://geocities.com/levelwater/shadow02.gif
https://web.archive.org/web/20080214044528im_/http://geocities.com/levelwater/siennaalejandria.gif

That is not even the main point.

The FE map relates to the RE globe mathematically as a projection.

There is no globe since you cannot justify how four trillion billion liters of water stay in place next to the outer surface of a sphere.

Take a look at this, half of the surface area of the globe is occupied by the Pacific Ocean:

https://images.fineartamerica.com/images-medium-large-5/satellite-view-of-earth-showing-the-pacific-ocean-copyright-tom-van-santgeosphere-project-santa-monicascience-photo-library.jpg

By contrast, the true flat earth Piri Reis map looks like this:

https://www.vhv.rs/dpng/d/449-4490274_theflatearthsociety-org-forum-sandokhan-piri-reis-map-hd.png

The surface area of the globe is four times as large as the surface area of the FE map.

How was it done?

https://web.archive.org/web/20210116142615/https://civilengineering-softstudies.com/2018/04/error-and-correction-for-the-curvature-of-the-earth-and-refraction-surveying.html

http://web.pdx.edu/~i1kc/courses/Surveying/Handouts/JGE%20Paper%201%20-%20Introduction.pdf (pg 14, curvature and refraction)

For the past 220 years, no matter which country, the CORRECTION for the supposed curvature of the Earth was added, automatically, for each and every surveying/topographical mission.

But here is something no one else has ever thought of: since there will always be some misalignment between the source and the reflector, the strings of light used by the theodolite will be subject to the influence of the Coriolis effect. That is, the path of the light strings will be deflected. This fact is not taken into consideration by land surveyors at all.

The surveying was done on land of course, and since they needed a huge surface area for the Pacific Ocean, they simply added it on maps with no problem.

Why is the radius of the flat earth so important? Because it tells us immediately when the next magnetic pole shift will take place. Since s = r x θ and we can't do anything about the angle [θ = 0.959582 rad (2 x 27.49 = 54.98, 54.98° = 0.959582 rad, 1/27.49 = 0.036376864 = 0.1 - 0.063623), Tropic to Tropic (solstice to solstice)], only the value of the radius is in question. I believe the radius of the FE measures 6,363.63 km.

The westward precessional shift of the Sun (1.5 km/year or 4.2 meters/day) must be taken into account.

The distances on the supposed globe have been increased greatly, see the distortions in Antarctica as an example:

http://www.moonglow.net/eclipse/2003nov23/

The photographers must be located some few hundred kilometers from the Sun (height of 636 meters), not the thousand of kilometers which would be assumed to be correct; given this fact, the Apollo 11 mission astronauts might as well have taken a sledge with reindeers to the Moon.





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Offline Tom Bishop

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Re: FE radius (UAFE estimate)
« Reply #3 on: August 17, 2023, 05:54:07 AM »
I think the point in this OP is that Erasthonese got the distance between the points wrong and so the Earth is smaller, under RE or FE assumptions. Maybe so, but the title of the article is Erasthonese on Diameter.

Erathoneses' original data should stay as primary focus, since the article is about his data and the interpretations that can be made from it. A subsection can perhaps be added that Uwe Topper thinks that he is wrong about the distances and the earth would be consequentially smaller with some details on what the FE size range would be under that scheme.
« Last Edit: August 17, 2023, 06:00:11 AM by Tom Bishop »

Re: FE radius (UAFE estimate)
« Reply #4 on: August 17, 2023, 06:11:17 AM »
There is no such thing as Erathosthenes original data, since he had used spherical triangles. The FE radius must be 6,363.63 km, upper bound less than 10,000 km. This is what we are talking about.

SteelyBob

Re: FE radius (UAFE estimate)
« Reply #5 on: August 17, 2023, 10:20:07 AM »


By contrast, the true flat earth Piri Reis map looks like this:

https://www.vhv.rs/dpng/d/449-4490274_theflatearthsociety-org-forum-sandokhan-piri-reis-map-hd.png


There are direct flights between New Zealand (Auckland) and Chile (Santiago). They take about 10.5 hours. I’m curious to understand where you think these flights would go if the world was indeed as presented on this map. Could you draw a line to show the route?

Re: FE radius (UAFE estimate)
« Reply #6 on: August 17, 2023, 11:26:42 AM »
The value of the radius is one of the most important elements of FET. If we know the radius, we can calculate the year as well as the month of the next reversal of the magnetic pole (the shift of the stellar dome as well).

Everyone on youtube, many other forums, is totally preoccupied with this issue, since it cannot be ignored anymore: when will the reversal of the magnetic poles take place? None of their dates can be justified, it is only FET which can offer a precise estimate.

You think that Auckland to Santiago de Chile was the most difficult route I had to deal with over the years? No, it was Juneau to Santiago de Chile:

https://www.theflatearthsociety.org/forum/index.php?topic=38712.msg961302#msg961302

« Last Edit: August 17, 2023, 11:40:05 AM by sandokhan »

Re: FE radius (UAFE estimate)
« Reply #7 on: August 17, 2023, 01:12:58 PM »


You think that Auckland to Santiago de Chile was the most difficult route I had to deal with over the years? No, it was Juneau to Santiago de Chile:

https://www.theflatearthsociety.org/forum/index.php?topic=38712.msg961302#msg961302


Interesting that you find these routes "difficult".  You didn't actually answer Bob's question about flying Aukland-Santiago did you? 

The reference you provided refers to a flight time Santiago-Sydney, and quotes a flight time of "18 hours .... perfectly compatible with my map".  I guess any flight time would be compatible, since the map does not actually include a scale of distances. 

Historical data on FlightRadar24 suggests that the flight times between the city-pairs Sydney/Santiago are typically 12.5 hours eastbound and 14 hours westbound (against the prevailing wind). 

Getting back to Bob's question, LAN801 is being operated today by Boeing Dreamliner CC-BGR; it left Santiago 7.51 hrs ago, and estimates landing Aukland in 4.09 hrs; exactly 12 hours.  Not only that, but the reciprocal flight, LAN800 is also currently airborne, reg CC-BGA, and estimates a flight time eastbound of 10.5 hours. 

Does all this fit your map? 

Re: FE radius (UAFE estimate)
« Reply #8 on: August 17, 2023, 01:43:14 PM »
The 12 hr flight would take place in the most direct way seen on the Piri Reis map. Is there a problem? Two possible flight paths: above Antarctica (as seen on the map) or passing between Antarctica and the Dome.

Here are two of the most unusual maps you'll ever see:





More information on the Piri Reis map:

https://www.bibliotecapleyades.net/mapas_pirireis/esp_mapaspirireis05.htm
https://www.bibliotecapleyades.net/mapas_pirireis/esp_mapaspirireis11.htm
https://www.bibliotecapleyades.net/mapas_pirireis/esp_mapaspirireis09.htm
https://stevedutch.net/pseudosc/1421.htm
https://stevedutch.net/pseudosc/piriries.htm

SteelyBob

Re: FE radius (UAFE estimate)
« Reply #9 on: August 17, 2023, 02:30:57 PM »
The 12 hr flight would take place in the most direct way seen on the Piri Reis map. Is there a problem? Two possible flight paths: above Antarctica (as seen on the map) or passing between Antarctica and the Dome.

Here are two of the most unusual maps you'll ever see:





More information on the Piri Reis map:

https://www.bibliotecapleyades.net/mapas_pirireis/esp_mapaspirireis05.htm
https://www.bibliotecapleyades.net/mapas_pirireis/esp_mapaspirireis11.htm
https://www.bibliotecapleyades.net/mapas_pirireis/esp_mapaspirireis09.htm
https://stevedutch.net/pseudosc/1421.htm
https://stevedutch.net/pseudosc/piriries.htm

Yes, there’s a problem. A straight line on your map goes past South Africa and then on to South America. And yet a direct flight from Sydney to Johannesburg, which on your map would involve going in a similar direction but nothing like as far, takes 14 hours or so. Why does it take longer to fly substantially less distance in the same direction?

Re: FE radius (UAFE estimate)
« Reply #10 on: August 17, 2023, 04:33:13 PM »


Look at the flight paths on a "globe" for Auckland-Santiago and Sydney-Johannesburg: on a FE map one of this routes would go between Antarctica and the Dome:







One flight path chooses to fly "above" Antarctica, while the other passes right between Antarctica and the Dome. Different durations of course.


Why did they choose the value of 6,363.63 km as the radius of the RE? Does it have anything to do some projection formula which would have worked if and only if 6,363.63 had been used?

Tesla had found out in 1908 that the map of Siberia was off by hundreds of kilometers. Most likely similar mistakes were involved in the distances attributed to Brazil, Argentina, Congo, Algeria, Lybia, Iran, Siberia, not to mention northern Canada.

SteelyBob

Re: FE radius (UAFE estimate)
« Reply #11 on: August 17, 2023, 05:01:31 PM »


Look at the flight paths on a "globe" for Auckland-Santiago and Sydney-Johannesburg: on a FE map one of this routes would go between Antarctica and the Dome:







One flight path chooses to fly "above" Antarctica, while the other passes right between Antarctica and the Dome. Different durations of course.


Why did they choose the value of 6,363.63 km as the radius of the RE? Does it have anything to do some projection formula which would have worked if and only if 6,363.63 had been used?

Tesla had found out in 1908 that the map of Siberia was off by hundreds of kilometers. Most likely similar mistakes were involved in the distances attributed to Brazil, Argentina, Congo, Algeria, Lybia, Iran, Siberia, not to mention northern Canada.

The flight paths on the globe are completely irrelevant. You have proposed that the earth is flat and that this particular map is the correct representation of it. My question was simply why, if that is the case, do the flight times not roughly match the apparent distances? Why does the shorter journey (Sydney - Jo'burg) take longer than a journey which appears to cover a much greater distance (Auckland - Santiago)?

Dual1ty

Re: FE radius (UAFE estimate)
« Reply #12 on: August 17, 2023, 05:31:48 PM »
Here's where the so-called radius of Earth comes from:

Of course, this "dome" is not physical because it is in fact the celestial sphere.


Re: FE radius (UAFE estimate)
« Reply #13 on: August 17, 2023, 05:39:36 PM »
Why does the shorter journey (Sydney - Jo'burg) take longer than a journey which appears to cover a much greater distance (Auckland - Santiago)?

Very simple: one journey is a straight line, while the other takes the airplane around Antarctica.


Now, we know for sure, absolutely 100%, that the radius of the Piri Reis map is 6,363.63 km, Charles Hapgood proved that fact while he had consulted with several noted mathematicians to help him out decipher the map.

Here is the question: we know that the Piri Reis map is an azimuthal equidistant projection (I believe the center is near the Marmara Sea, since the northern portion of the map is missing, but that is another matter). What, then, is the radius if the map would be projected onto a sphere?

Re: FE radius (UAFE estimate)
« Reply #14 on: August 17, 2023, 07:25:01 PM »
The answer to the question is related to the circumference of the Earth (Flat or Globe): it must give the same value in both versions.

https://archive.org/details/HapgoodCharlesHutchinsMapsOfTheAncientSeaKings/page/n27/mode/1up?view=theater (page 33 - the relationship between the value attributed to Erathosthenes and the radius of Piri Reis' map)

FE radius = 6,363.63 km

SteelyBob

Re: FE radius (UAFE estimate)
« Reply #15 on: August 17, 2023, 07:58:12 PM »
Why does the shorter journey (Sydney - Jo'burg) take longer than a journey which appears to cover a much greater distance (Auckland - Santiago)?

Very simple: one journey is a straight line, while the other takes the airplane around Antarctica.

Can you draw the approximate route you are suggesting that the flights take on your FE map? I’m not clear what you’re suggesting.


Re: FE radius (UAFE estimate)
« Reply #16 on: August 23, 2023, 06:52:02 PM »
Why does the shorter journey (Sydney - Jo'burg) take longer than a journey which appears to cover a much greater distance (Auckland - Santiago)?

Google maps says Sydney - Jo'burn is about 11,000 kms, and Auckland - Santiago, Chile is about 9,700 kms, so your premise is based on an error. Where are you getting your distances from?

Re: FE radius (UAFE estimate)
« Reply #17 on: August 28, 2023, 04:47:59 AM »
The value of the radius is one of the most important elements of FET. If we know the radius, we can calculate the year as well as the month of the next reversal of the magnetic pole (the shift of the stellar dome as well).

Everyone on youtube, many other forums, is totally preoccupied with this issue, since it cannot be ignored anymore: when will the reversal of the magnetic poles take place? None of their dates can be justified, it is only FET which can offer a precise estimate.

You think that Auckland to Santiago de Chile was the most difficult route I had to deal with over the years? No, it was Juneau to Santiago de Chile:

https://www.theflatearthsociety.org/forum/index.php?topic=38712.msg961302#msg961302

How would you calculate the FE radius if you guys don't know the extents of FE into Antarctica?  The azimuthal Gleason projection has a 12,500 mile radius because it has a defined south "edge."