#### poweraide

• 6
##### Re: Found a fully working flat earth model?
« Reply #120 on: February 05, 2022, 08:50:38 PM »
Can they? Because on a flat earth, they'd have to reflect off something. There's no connection between one edge and the other, correct?

And once waves reflect, they won't return to the same spot, unless the wave starts from the very center of the transformation.
« Last Edit: February 05, 2022, 08:58:06 PM by poweraide »

#### stack

• 3583
##### Re: Found a fully working flat earth model?
« Reply #121 on: February 05, 2022, 09:04:23 PM »
I'm not sure this is entirely relevant, but it kind of looks similar to the creation of the model. Troolon, you may (or may not) find this interesting:

Transforming from Geographic to Celestial Coordinates

The simplest astronomical observation of all is that the stars appear to move around the Earth (which, of course is the result of the Earth rotating with respect to the ‘fixed’1 stars). We therefore catalogue the position of astronomical bodies against a ‘Celestial’ coordinate system that is fixed with respect to the stars, but moves around with respect to Earth based coordinates.

The plan will involve four stages:
1. Convert to a cartesian (‘xyz’) coordinate system, with x pointing north, y pointing west, and z to the sky. (Y to the west gives us a right-handed coordinate system, which is the same as the celestial coordinate system.)
2. Rotate the axes of our coordinate system around the east-west direction so that the z axis is now parallel with the rotation axis of the Earth.
3. Rotate the axes around the new z direction (we will call this z 0) until the x axis points towards the vernal equinox, which is the point on the sky where in its annual progress around the ecliptic plane the Sun crosses the celestial equator moving northwards (usually about 20th of March).
4. Convert back from cartesian to polar coordinates to get the celestial values.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #122 on: February 05, 2022, 10:20:20 PM »
I'm not sure this is entirely relevant, but it kind of looks similar to the creation of the model. Troolon, you may (or may not) find this interesting:

Transforming from Geographic to Celestial Coordinates
Awesome, thanks!
I was already looking for a way to convert RA/dec to lat/long at a given time. If only to make the simulation more realistic.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #123 on: February 05, 2022, 10:26:27 PM »
Can they? Because on a flat earth, they'd have to reflect off something. There's no connection between one edge and the other, correct?

And once waves reflect, they won't return to the same spot, unless the wave starts from the very center of the transformation.
The boundary of the disc is nothing but a mathematical artifact. Things crossing the southpole, appear at the other end of the map.

But as a general answer, my 3D model is globe physics, just rendered differently. If you take a globe, express it in in celestial coords (lat, long, distance), and then draw latitude on a straight instead of a radial axis, you'll get a flat-earth-universe. Physics works with celestial coords, and so it behaves like a globe, it just looks flat. So if a wave works on a globe, it also works on the flat renderering. The only difference is the shape of one axis.

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #124 on: February 06, 2022, 07:51:22 AM »
Quote
flat earth can not work with cartesian distances
This is exactly what you are not getting.  There is no such thing as “cartesian distance” or “spherical distance” or “polar distance”.  When you transform the metric tensor correctly, the distances are the same no matter what coordinate system you use
Quote
The distance between two points will not change if you change your coordinates. Also the metric tensor itself, as a function, will not change. But the matrix which represents the metric tensor depends on what coordinate system is being imposed on all of the tangent spaces, so the matrix representation of the metric tensor will in fact change with changes of coordinates.
The whole point of coordinate transformations is that you don’t change the physical nature or geometry of the underlying structure.  If your transformation results in different distances between points and you have to use a "bendy ruler" to make the distances work, that is exactly what you have done.
« Last Edit: February 06, 2022, 07:55:05 AM by Rog »

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #125 on: February 06, 2022, 04:51:11 PM »
the distances are the same no matter what coordinate system you use
agreed
Quote
The whole point of coordinate transformations is that you don’t change the physical nature or geometry of the underlying structure.  If your transformation results in different distances between points and you have to use a "bendy ruler" to make the distances work, that is exactly what you have done.
agreed again.
The width of Australia should be around 4000km whether calculated in cartesian or celestial coords.
I believe the only discussion we are still having is that i'm claiming the picture below is a sphere expressed in celestial coords, whereas you believe it's a disc expressed in cartesian coords.
Just looking at the width of Australia, we can be pretty sure, cartesian coords are incorrect. When looked at in celestial coords, distances do match with the globe model.

Mathematically the above picture is a sphere. Just like the 2 graphs below are both representations of the same sine function yet look totally different. Math only cares about numbers, not shapes.

And this brings us back to my original point: physics works regardless of shape. It only cares the numbers and formulas are correct and surprising enough, the numbers work on different shapes... ergo  the shape of the planet can never be known.

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #126 on: February 06, 2022, 08:35:41 PM »
You continue to miss the point.  If you did the transformation correctly, the distance would be the same in any coordinate system

Quote
Mathematically the above picture is a sphere. Just like the 2 graphs below are both representations of the same sine function yet look totally different

Just because the coordinates are the same, doesn't mean that mathematically it is a sphere.  A sphere has the least amount of surface area per volume of any shape.  All you've done is manipulate the amount of surface area to fit the same volume.

http://walter.bislins.ch/bloge/index.asp?page=Globe+and+Flat+Earth+Transformations+and+Mappings&q=coordinate+transformation#H_WGS84_Coordinate_System
Example above: Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same. They have different directions and lengths and hence different cartesian coordinates. The lengths of the magenta vectors are shown at |Vglobe| and |Vfe| respectively [/quote]

The vectors represent a measurement of area.  So even though the coordinates are the same, the coordinates represent a different measurement of area. That’s exactly what is not supposed to happen when you transform coordinates from one system to another.  If you transform the metric tensor correctly, the coordinates in any system should represent the same amount of area in any other system.

Quote
Note: Coordinate System Transformations (CS Transf) do not change the length and direction of vectors. They only change the values of the vector components to get the same vector in the corresponding coordinate system.

Translation: A correctly performed transformation doesn't change the amount of surface area.

.Sure, you can manipulate the same volume of space of a sphere and plot it on a disc, but you have to ignore the rules about transforming the metric tensor and therefore aren’t represent reality. You can change the shape of anything to look like anything, but if you ignore the rules of physics doing so, it isn’t much of an accomplishment.

Quote
And this brings us back to my original point: physics works regardless of shape.

Not if you follow the rules of physics to determine the shape.

Also, another question.  What mechanism for gravity are you assuming?  If you are sticking with UA, then you should probably be using Rindler coordinates.
« Last Edit: February 06, 2022, 08:50:23 PM by Rog »

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #127 on: February 06, 2022, 09:13:45 PM »
You continue to miss the point.
At least one of us does
Can you please answer the question i've asked multiple times before. At what point do you believe the transformation breaks?
1. When i'm switching from cartesian to celestial coords,
2. or when i'm drawing latitude on a straight axis instead of a curved axis?

Quote
Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same.
Quote
FYI: The link only contains black/white pictures. I did not find magenta vectors

Quote
Just because the coordinates are the same, doesn't mean that mathematically it is a sphere.  A sphere has the least amount of surface area per volume of any shape.  ...
Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same. They have different directions and lengths and hence different cartesian coordinates. The lengths of the magenta vectors are shown at |Vglobe| and |Vfe| respectively
But my vectors are the same...  A point in celestial coords can be expressed in terms of its vectors as:  x*latitude-vector + y*longitude_vector + z*distance-vector.
(note i'm using vector very liberal here as latitude and longitude are a circular and not straight axis. It's probably more logical to express a point as distance along the equator plus distance along the northpole-southpole-greatcircle)
On the flat cylinder, the same point is still expressed as: x*latitude-vector + y*longitude_vector + z*distance-vector.

So as far as the math is concerned, these 2 points are the same. Same values, same vectors

Quote
The vectors represent a measurement of area.  So even though the coordinates are the same, the coordinates represent a different measurement of area. That’s exactly what is not supposed to happen when you transform coordinates from one system to another.  If you transform the metric tensor correctly, the coordinates in any system should represent the same amount of area in any other system.
area is the same as axis and coordinates of any point are the same.
eg: in celestial coords, the sphere with equation lat=[0-π], long=[0-2π], distance<=R  has volume 4πR²
The flat cylinder with equation  lat=[0-π], long=[0-2π], distance<=R  also has volume 4πR²
There is no difference.

Quote
Note: Coordinate System Transformations (CS Transf) do not change the length and direction of vectors. They only change the values of the vector components to get the same vector in the corresponding coordinate system.
Translation: A correctly performed transformation doesn't change the amount of surface area.
Sure, you can manipulate the same volume of space of a sphere and plot it on a disc, but you have to ignore the rules about transforming the metric tensor and therefore aren’t represent reality. You can change the shape of anything to look like anything, but if you ignore the rules of physics doing so, it isn’t much of an accomplishment.
I'm not ignoring any rules. In fact i'm relying on all these rules not to break the physics. If distances or vectors changed, everything would break.
It's just a different representation, but it are celesitial coordinates. You should calculate distances, angles and areas, with to celestial coordinate formulas.

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #128 on: February 06, 2022, 09:23:51 PM »

"Intuition: Fill space with concentric sphere’s around the origin. For every sphere , transform it into a disc with an an azimuthal projection and finally insert the disc into a cylinder at a height equal to the radius of the sphere.
transforming a sphere"

You are not "transforming a sphere", you are transforming 3space. The transformation includes flattening each sphere, which is not part of the process of coordinate conversion. Because we are only interested in the earth, we can ignore all of the stack of disks except the one with the radius of the earth. You can shorten the statement to "take the earth and transform it into a disc". This is not changing coordinate systems, this is just flattening the disk, just like the maps in the FAQ and with the same distortion.

If you represent the same sphere in different coordinate systems and use the equations for that coordinate system, you will get the same distances, shape, size in all coordinate systems. To convert a sphere in cartessian to spherical, find the instructions here:

https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_12%3A_Vectors_and_the_Geometry_of_Space/12.7%3A_Cylindrical_and_Spherical_Coordinates

In all coordinate systems, you will end up with a sphere where every point is equidistant from the origin. Your AE/FE projection does not do that, so not a sphere.

On a spherical earth, longitude lines below the equator converge, get closer together. On disk earth (FE/AE) they diverge. Case in point, the coasts of Australia. If we take their longitude as the same on RE and FE, on RE the longitude lines are closer than the equator, less distant. On FE, they diverge, the lines are farther apart, more distance. Distance is not preserved, not equivalent. This is what happens when you "straighten the longitude lines". Distance is distorted. The appearance of the AE confirms that Australia is way too wide. We can do the calculations of what the distance would be with converging longitudinal lines and diverging. Only one can match, the other will be falsified.

Still wondering where Sigma Octantus is. Do I understand your reply to be that you can make a graphic of Sigma Octantus light bending however it needs to so that everyone in the southern hemisphere sees it directly south at an angle of inclination equal to their latitude? Perhaps, like the shape of the earth, no one can ever know? Seems like a pretty weird coincidence that the light would bend however it needs to to make Polaris be directly north at an angle of inclination equal to your latitude.

Do you agree that RE geometry explains this with straight light and a reasonable location for Sigma Octantus? That on FE, no one knows where Sigma Octantus is or how or why the light bends?

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.

#### spaceman spiff

• 24
##### Re: Found a fully working flat earth model?
« Reply #129 on: February 07, 2022, 01:27:27 AM »
I don't get what all the fuss is about, really, unless I've missed something.

Quote from: troolon
But as a general answer, my 3D model is globe physics, just rendered differently. If you take a globe, express it in in celestial coords (lat, long, distance), and then draw latitude on a straight instead of a radial axis, you'll get a flat-earth-universe. Physics works with celestial coords, and so it behaves like a globe, it just looks flat. So if a wave works on a globe, it also works on the flat renderering. The only difference is the shape of one axis.
The OP him/herself said that the model is globe earth rendered on a flat disk, with whatever it takes to make the shape of the sun's illumination match the observations, and calling the boundary of the disk a  "mathematical artefact".  It is not flat earth. The title of the thread is a bit disingenuous.

Saying that you can't differentiate between the models is also strange, since there aren't two models, but only one represented in different ways.

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #130 on: February 07, 2022, 02:13:30 AM »
The OP is claiming that the distances and physics of his model work the same as RE. This is not true, a disk is not equivalent to sphere, the distances are different. What he claimed was changing coordinates included flattening a sphere into a disk. Light has to bend in ways that there is no evidence of. He claims the earth can be any shape, physics works the same, they are mathematically equivalent. This is not true.

You can't prove the earth is round, but you can do 2 things:

1. Assume the earth is flat and explore the implications. You can observe that distances on the AE map are wrong, that there is no good explanation for some people seeing the dome as daylight while at the same time others see it as night. In the northern hemisphere, people see the north star directly north at an inclination equal to your latitude. This works on AE/FE for left/right (azimuth), but there is no place where it works for up/down (altitude) without bending the light. This is called Electromagnetic Acceleration in the FAQ, which the FAQ says is "unknown forces with unknown equations". In the southern hemisphere, everyone can see the southern pole star directly south, which means outward from the globe, thus in Australia and South Africa directly opposite directions at the same time.

This means either the laws of physics are wrong and measurement has to be flexible - he says FE requires bendy rulers, and he is right, but that's not measurement, it is adjusting your observations to fit. This falsifies the model.

2. You can observe that RE explains this quite reasonably with light traveling straight and no contradictions with observations.

The OP made several errors of thought, thinking he was doing a coordinate conversion when he was transforming the object. thinking that the disk he changed it to was a sphere when looking at it it is clearly a flat circle, using spherical formulas on 2d polar coordinate object, thinking that transforming the shape would not change the physics.

RET explains this and much more, OP can't explain where Sigma Octantus is. He can't show day/night sky on the dome, where half the world sees light blue daylight and half see darkness with stars over the entire dome at the same time.

If he wants to say it is a thought experiment illustrating an impossible world, well, that's cute. If he wants to say his model is valid and a possible alternative to RE with equivalent physics, he is wrong.

Again, his conversion flattens the globe, while he claims he is doing a coordinate conversion. It is not the same geometric shape. A sphere is not a disk. You can map the surface of a sphere onto a disk, but that is not the same model.
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.

#### scomato

• 175
##### Re: Found a fully working flat earth model?
« Reply #131 on: February 07, 2022, 02:54:23 AM »
If OP is claiming that via arbitrary transformation the globe can be a disc and vice versa, what’s to say then the Earth isn’t a rhombus, donut, non-euclidean, or shaped exactly like Howard Stern?

If we ignore reality we can arbitrarily transform one coordinate system into another, welcome to the history of map making, this was never even a point of debate in the first place and I fail to see how it has any relevance to Flat Earth theory at all.

#### Clyde Frog

• 1045
• [kʰlaɪ̯d fɹɒg]
##### Re: Found a fully working flat earth model?
« Reply #132 on: February 07, 2022, 02:17:56 PM »
If OP is claiming that via arbitrary transformation the globe can be a disc and vice versa, what’s to say then the Earth isn’t a rhombus, donut, non-euclidean, or shaped exactly like Howard Stern?
If you had read basically anything OP posted in this thread you are posting in right now, you'd know they said exactly this already.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #133 on: February 08, 2022, 10:12:34 PM »
...
Because we are only interested in the earth, we can ignore all of the stack of disks except the one with the radius of the earth. You can shorten the statement to "take the earth and transform it into a disc". This is not changing coordinate systems, this is just flattening the disk, just like the maps in the FAQ and with the same distortion.
Why would a flattening not be a coordinate transformation? You can go back and forth between them, no information loss. The surface of a sphere is a 2D structure, so it a disc. Here's the transform to go back and forth between both:
(lat, lon, dist) = (asin(z / √(x²+y²+z²)), atan2(y, x), √(x²+y²+z²))
(x,y,z) = (dist * cos(lat) * cos(lon), dist * cos(lat) * sin(lon), dist *sin(lat))

Quote
If you represent the same sphere in different coordinate systems and use the equations for that coordinate system, you will get the same distances, shape, size in all coordinate systems. To convert a sphere in cartessian to spherical, find the instructions here:
In all coordinate systems, you will end up with a sphere where every point is equidistant from the origin. Your AE/FE projection does not do that, so not a sphere.
Please remember that on a sphere in spherical coordinates, coordinates are expressed as (lat, lon). Or degrees along the equator-circle+ degrees along the NS-circle. In the flat rendering points are still expressed as degrees along the equator-circle+ degrees along the NS-line
I know this sounds counterintuitive as we use cartesian coordinates so often, but on the cylinder, the 3 axis are the equator, NS-line and height. Any point is expressed as a "sum of these 3 axis".  When using these 3 axis, all points are indeed equidistant from the center of the earth.
The easiest way to show is:   center of the earth in spherical coords = (0,0,0). Any point on the flat earth has coordinates([0-π], [0-2π], 6000km). The distance between any point and the origin is exactly 6000km. Exactly as expected.
Do note that if we represent coordinates in cartesian form (x,y,z) and then calculate the distance, everything will be broken. You must respect the axis.

Quote
On a spherical earth, longitude lines below the equator converge, get closer together. On disk earth (FE/AE) they diverge. Case in point, the coasts of Australia. If we take their longitude as the same on RE and FE, on RE the longitude lines are closer than the equator, less distant. On FE, they diverge, the lines are farther apart, more distance. Distance is not preserved, not equivalent. This is what happens when you "straighten the longitude lines". Distance is distorted. The appearance of the AE confirms that Australia is way too wide. We can do the calculations of what the distance would be with converging longitudinal lines and diverging. Only one can match, the other will be falsified.
Remember our axis are the "equator" and NS-line. On a sphere, points are expressed as (lat/lon) -- degrees along the equator + degrees on the NS-circle. On the flat earth they are too. When you express coords as lat/lon, you should use the lat/lon distance formulas (ie haversine) and Australia will have the correct width also on the flat earth.
To give a quick example. the left-most point of australia on the flat map has coord (-33°, 115°). The rightmost point (-29°, 154°). Calculating the distance gives: 3722km (which matches google maps and the globe model)
Calculating the distance with xy-coordinates will indeed give horribly deformed results.

Quote
Still wondering where Sigma Octantus is. Do I understand your reply to be that you can make a graphic of Sigma Octantus light bending however it needs to so that everyone in the southern hemisphere sees it directly south at an angle of inclination equal to their latitude? Perhaps, like the shape of the earth, no one can ever know? Seems like a pretty weird coincidence that the light would bend however it needs to to make Polaris be directly north at an angle of inclination equal to your latitude.
If you had asked for 3 people looking north and seeing polaris, the answer would be really straightforward on the disc. Draw a star above the northpole, draw curvy rays from the 3 observers going to the star and done.
To understand the southstar, there's one peculiarity about the AE map you must understand: The southpole, a single point on the globe, gets transformed into a circle. The entire outer edge of the AE map represents just 1 point in reality. Mathematically: the southpole has coordinate (lat,lon) = (-90°, [0-2π]) There is no unique coordinate for this point.
The same phenomenon happens with points directly below the southpole: it gets transformed into a circle. So the southstar in the flat earth universe is rendered as a circle rather than a point. It's rather obvious that 3 people looking south, can see the same circle.
If we move the southstar just the slightest bit away from the southpole, it would again become a single point. When you then draw the rays of light, you will see they start due south from the observer, and then start curving all around the disc to meet up with the southstar. I can promise you, mathematically it is all correct even though it's not very intuitive.

Quote
Do you agree that RE geometry explains this with straight light and a reasonable location for Sigma Octantus? That on FE, no one knows where Sigma Octantus is or how or why the light bends?
Re explains the world very well.
On FE, sigma octantis position is given by (lat, lon, distance). We know perfectly well where it is.
The reason light bends in the FE rendering is the same reason light travels straight in RE.
Remember both RE and FE share the same physics. There's not difference. It's just a different representation.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #134 on: February 08, 2022, 10:22:59 PM »
I don't get what all the fuss is about, really, unless I've missed something.
Based on your reactin, you have probably totally understood.
From a physics viewpoint this model has nothing new to offer.
However in the context of the flat-earth-"debate", it offers a few interesting tidbits:
- There exists a working flat-earth model. People have been looking for this for over 100 years
- The true shape of the earth can never be known. It can be a globe, or flat or a pyramid. There's no way to measure it. It seems people do not always realize this.

But yes, you have probably understood everything correctly and your reaction is very typical for a physics graduate.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #135 on: February 08, 2022, 10:29:23 PM »
If OP is claiming that via arbitrary transformation the globe can be a disc and vice versa, what’s to say then the Earth isn’t a rhombus, donut, non-euclidean, or shaped exactly like Howard Stern?
I totally agree. Physics can be made to work on any shape planet.

Quote
If we ignore reality we can arbitrarily transform one coordinate system into another, welcome to the history of map making, this was never even a point of debate in the first place and I fail to see how it has any relevance to Flat Earth theory at all.
We're not ignoring reality. Physics works regardless of shape. You can make physics work equally well on a globe, a disc or Howard Stern.
However as all these shapes produce the same answer to any calculation, it's impossible to differentiate the models.
And so in reality there's no way to measure if we're on a globe, or a disc. We could be on a globe with straight light, or we could be on a disc with bendy light.
There's no measurement or observation to find out.
And in my mind, this pretty much settles the entire flat-earth-"debate". Any shape earth is possible and we'll never be able to find out what the real shape is...

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #136 on: February 08, 2022, 10:58:01 PM »
The OP is claiming that the distances and physics of his model work the same as RE.
<paraphrased> but nothing matches </paraphrased>
I've given you the math.
I've given you examples.
I've told you how you should calculate distances.
I've created loads of images and animations: day, night, seasons, southstar, ...
I've explained how to create the images yourself.

The graphics and animations match expectations in reality (ie day/night/seasons/keels of ships disappearing, ...)
Quite a few people in this forum have understood.
I've had this work validated by physicists prior to posting.
I've managed to explain a computer to do all of this.
I'm currently working on adding all planets in the solar system at their correct positions.
And yet you keep coming back to the same questions that if true would break all of the above results.
I'm not sure anymore how i can explain it any better.

This model is globe physics. Mathematically there is no difference. Yes it looks flat, but to mathematics it behaves as a globe.
Mathematics doesn't care about shape. It can't see what you've drawn on your piece of paper.
The reason it fails your intuition is because there is no XYZ axis. Coordinates are expressed as lat/long: degrees along the equator + degrees along the NS-line.
In this axis everything makes sense.

#### stack

• 3583
##### Re: Found a fully working flat earth model?
« Reply #137 on: February 08, 2022, 11:27:17 PM »
Troolon, you might find this site interesting. It allows you to custom generate a couple of dozen different map projections. I'm not sure what transformations it uses, but it may be the same as yours.

WORLDMAP­GENERATOR

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #138 on: February 09, 2022, 07:09:08 AM »
You broke physics (actually geometry is probably more accurate) when you decided to just make up your own rules for a metric. I’ve wracked my brain trying to figure out a way to explain to you, in a way that you will understand. It is difficult because you don’t seem to grasp the very basic concept that Euclidean (flat) and non-Euclidean geometry (spherical) do not follow the same rules.  Most importantly, distances are measured differently

In differential geometry, distances are measured by a metric.  Euclidean and non-Euclidean spaces use different metrics. They have to because the definition of  “the shortest distance between two points is different”.  For Euclidean space it is a straight line, for non-Euclidean it is a curve. For Euclidean space the metric is simply the Pythagorean theorem.  In non-Euclidean space it is

Obviously, very different formulas

When you do a coordinate transformation between two Euclidean spaces, the metric remains the same so lengths and angles are preserved.  When you do a coordinate transformation from a Euclidean space to a non-Euclidean space (Cartesian on a flat disc to Celestial on a sphere, for example), you aren’t just transforming the coordinate system, you are also transforming the metric from Euclidean to non-Euclidean. You seem to have just arbitrarily transformed the metric by “updating the formula”.  You can’t do that. The metric transforms when you do the coordinate transformation. You can’t just decide that you don’t like the way it transforms and invent your own metric.

But even when done correctly, because the bases of the two different metrics (the underlying geometry) are different, it will never transform exactly. Non-Euclidean metric defines distances in terms of angles, Euclidean doesn’t.

#### spaceman spiff

• 24
##### Re: Found a fully working flat earth model?
« Reply #139 on: February 09, 2022, 03:21:15 PM »
I don't get what all the fuss is about, really, unless I've missed something.
Based on your reactin, you have probably totally understood.
From a physics viewpoint this model has nothing new to offer.
However in the context of the flat-earth-"debate", it offers a few interesting tidbits:
- There exists a working flat-earth model. People have been looking for this for over 100 years
- The true shape of the earth can never be known. It can be a globe, or flat or a pyramid. There's no way to measure it. It seems people do not always realize this.

But yes, you have probably understood everything correctly and your reaction is very typical for a physics graduate.

First bold quote contradicts what you said. You don't have a flat earth model, you have a globe earth model represented in a disk. A projection, nothing new. Then you make everything else fit to that shape, starting from a globe.

What people are trying to say is that reality gets in the way. If your model is meant to represent a truly, real, physical flat earth, there is absolutely no reason whatsoever for the scale of the map to be different depending on the location, and the distance metric should be euclidian everywhere. This clearly does not work, as you have mentioned about Australia.

On the second point, it can only be right if you disregard physics completely and only care about a purely mathematical model. The shape of sunlight, sunsets and sunrises, constellation retaining shapes throughout the night... all these are observations that point to a globe. You could MAKE them fit any projection of the earth you want, sure. But the instant you claim that any projection is actually reality, you'll need to invoke completely new and unsupported (by evidence) physics to explain the behaviour of light, the fact that people circumnavigate Antarctica, observations of earth from space, etc.

And to finish it, if you calculate the curvature of your model to be different from 0, you don't have a flat earth. I hope this is clear