#### Tumeni

• 3179
##### Re: Found a fully working flat earth model?
« Reply #160 on: February 13, 2022, 11:21:13 AM »
" i had it scrutinized by graduated physicists, knowledgeable in the field, prior to making it public. They called it "logically sound, but practically useless"

"in general, it's mathematically impossible to find a difference as flat and globe are the same model."

Doesn't this suggest that practically, there is, or may be, a difference?

=============================
Not Flat. Happy to prove this, if you ask me.
=============================

Nearly all flat earthers agree the earth is not a globe.

Nearly?

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #161 on: February 13, 2022, 12:03:46 PM »
Carpeting Australia
first off: very vivid example. I love you approach.

How would you create an Australia sized carpet using the globe model? I would imagine it goes something like this:
- with a ruler measure the width and height of Australia on a scale model.
Please note that this is a bit tricky as rulers don't fit to globes very well, and measuring straight angles is also going to require quite a bit of finessing.
- convert all measurements according to the model. As you've been measuring scaled distances, converting this to a real world carpet will require some math. Doubly so if you want to express all properties of the shape (like curvature).

The process for the flat-earth representation is completely similar:
- with a flat-earth ruler measure the width and height of Australia (do note that our axis is not translation invariant and we measure lat/long in degrees, not in cartesian kilometers)
- convert all measurements according to the model. We measured degrees lat/lon, so convert to kms using the regular formulas. Note that this conversion turns flat areas of disc, back into areas of a sphere (in an orthonormal basis speaking of course).

So the same process creates the same real-world carpet in both models. What you of course can't do, is measure on the flat-reath represenation with a globe ruler, or convert using globe-formulas.

Question 1; Hobart is around 42 deg S latitude.  At what point, if at all, should our technician be using bendy rulers or variable-scales to draw his map, because the drawing is, after all, a map?
[/quote]
First off, for observers inside the model all measurements fit with reality (and the globe model).
So the technician can use a straight ruler on site. 3mx4m, 5m diagonal will be correct in London and Australia.
For making a map, you should start deforming once the distortion becomes too big to become a factor.
(just like you would switch from flat to spherical models once the curvature starts to matter)

Quote
<paraphrased>Create a carpet for all of Australia: 3000km x 4000km, and extend a little over the coast.  So we manufacture a carpet loom 3000km wide, and use it to produce a carpet 4000km long.
So now the Aussies roll it out from Brisbane towards the West.
How does it look?  Is it wrinkled at the edges?  Is there a ridge over Melbourne? If we push it down over Victoria, does it leave a gap over Sydney?
It will be wrinkled.
When expressed in an orthonormal coordinate system: only a spherical carpet will fit Australia.
When expressed in my flat-earth coordinate system, the carpet will have all the spherical properties for math and any observer, but when you were to place it on the flat-earth representation, it would look flat (just like the sphere looks flat)
If you've read my pie-chart versus bar-chart analogy, what you're doing is taking a pie out of a pie-chart and trying to fit it onto a bar-chart. You need to transform the pie to a bar first, then it will work.

Quote
Perhaps it lies perfectly flat; no gaps.  We have proven that the Earth is flat, or cylindrical, or a cone, or a pyramid or some-such.  We possibly don't know which, but we have completely refuted the idea that the world is spherical, or shaped like Angela Merkel.  If we continue to identify what shapes it isn't, eventually we will know what shape it is.
You can't take a square in an orthonormal axis and cut and paste it onto a logarithmic scales without changing it's meaning. In the flat model we use lat/long coordinates, not cartesian ones. You can't copy/paste cartesian shapes onto it and expect it to make sense. This particular model will transform spheres into cylinders and so a flat Australia in the model, really is a spherical area according to the math and observers living inside the model.

Quote
Your contention that we cannot know the shape of anything is a nonsense.  Its not just about the maths.  The Earth, like our fictional carpet, is an entity with fixed dimensions.  If we measure it enough ways, using a constant metric against entities who's size we know, we can know its shape or, at the very least, identify what shape it isn't.
A sphere is just a representation of reality. A representation in an orthonormal axis.
A cylinder is another representation of reality.
It's like bar-charts versus pie-charts. They're just representations. You can't infer the true shape from a mere representation.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #162 on: February 13, 2022, 12:30:57 PM »
Fine, but we live in the warehouse.  We can walk around it, climb it, measure it.  Its a warehouse.
This is why i explicitly stated our only provided interface is the provided computer interface.
But fine, let's explore this new example a little further:

The problem is that you can only measure things with things inside the warehouse.
If i made the warehouse twice as big, you and your ruler included, could you tell?
If i rotated the warehouse 90 degrees, how would you know?

In the real world you're always measuring 2 quantities at the same time.
- curvature of the earth versus curvature of light (ie laser of over lake)
- distances on the ground versus shape of your ruler/coordinate systems
- rotation of earth versus rotation of space
- ...
We can only ever make relative measurements.
And mathematically it can be shown that you can remove the curvature of earth whilst changing everything else in the universe to compensate. You wouldn't be able to tell.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #163 on: February 13, 2022, 01:44:20 PM »
" i had it scrutinized by graduated physicists, knowledgeable in the field, prior to making it public. They called it "logically sound, but practically useless"

"in general, it's mathematically impossible to find a difference as flat and globe are the same model."

Doesn't this suggest that practically, there is, or may be, a difference?
In general the math is more difficult, but it will eventually yield the same result.
Remember how in maths you switch to polar coords because it's sometimes easier to calculate with them?
Quick example: spherical distance in lat/long coordinates is the haversine formula. In cartesian coords it's the double differential Rog showed (or a path-length integral). Result will be the same but haversine is probably easier to work with.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #164 on: February 13, 2022, 02:09:33 PM »
Occam's razor is not what is most useful, it says the simplest explanation is the best.
Occams Razor is a philosophical idea, not a law of nature. It does not invalidate any model.

That said, there are various interpretations, and simplest explanation or least assumptions interpretation i find really tricky.
Assuming magic or divine intervention is only one assumption...
When i look in practice, physics has at least 5 alternative string theories for particle physics and they're all useful.
You have newtonian gravity and it's alternative model: relativity.
In practice you need to understand the benefits and limitations of your model and go with the most useful one for your problem.
But this is now philosophy and not science. So i'm perfectly happy that you have your own interpretation.

One small remark though, i believe the number of assumptions between both models is the same:
- why does light bend <-> why does light travel straight
- why does the earth rotate <-> why do the stars rotate around earth.
- cartesian coords <-> celestial coords
...
So the number of assumptions are imo equal and i find that interpretation of Occam's razor hard to apply.

Quote
As for your notion that we can't be sure about things, that is called solipsism, and has a philosophical history. You can hold the position that nothing can be known except that you exist to ask the question.
I'm no expert in philosophy and this is definitely related to solipsism. However i believe there to be a crucial distance: I assume there is a universe and a planet and  light and that it all behaves in predictable patterns. I believe we can create laws that describe relations between objects, but we can't ever know the true shape universe.

Quote
Gps works by sending a timestamp to your device, where the local time is subtracted, and the speed of light is used to calculate your distance from the satellite, a form of LIDAR. If the light is bending all over the place and we don't know where the satellites are, how does this work? All RET, deniable only with conspiracy, changed laws of physics, and denial of known science.
The same as in the globe model. Remeber it's just a different representation. I'm not denying or invalidating RE. In fact i rely upon it.

Quote
Can I request that you make another graphic? Pick 10 cities around the globe (more is better) and get their distances with google maps or airline schedules. Take a point at the center of your graphic and plot a point 3963 miles from it (radius of earth). Plot a second city also 3963 from origin and at the end of an arc with center at origin and arc length equal to the distance between each city. Continue plotting cities and their connecting arcs. If the result is a sphere the earth is round. If not, it is not round.
If i find time but i won't promise anything. I can already tell you the way the code works is by transforming circles back and forth to lat/long and you'll just end up with a bunch of deformed circles on a map. Mathematically it will make sense. BTW, to see a circle in my model, look at the animations. The yellow dots are the dawn/dusk areas and on the globe they're a great-circle on the sphere.

Quote
If you get on a plane to Hawaii, do you want the navigation to be by RET or FET?
Don't care. The result is the same, they're indistinguishable. It's just a different representation.

Quote
I don't know who you talked to about your math, but measurement is measurement, the same everywhere or meaningless, and changing coordinate systems in Euclidean (the world we live in) changes nothing about the math, only the notation, not the size or shape.
Then please take a piece of paper and draw a logaritmic X and Y axis. Now please draw a circle. (for example all points at distance 1000 from a center at (1000,1000)).  The circle will be deformed. Have fun measuring that with your regular ruler

#### stack

• 3583
##### Re: Found a fully working flat earth model?
« Reply #165 on: February 14, 2022, 08:57:35 PM »
Bendy light matches observation.
Shine a laser beam over a lake. You measure curvature. This curvature can be either explained by the earth bending or light bending. Impossible to tell.
There is no test to differentiate the shapes.

What if you don't use a light/laser to survey? Like maybe a Theodolite.
I'm no expert on theodolites, but i believe they're still sight based.
But even in general, it's mathematically impossible to find a difference as flat and globe are the same model.

You might find this interesting regarding lasers over distance. Specifically how the 4KM long arms of the two LIGO sites were measured and constructed creating a straight light path in relation to Earth's curvature. There are some transformations they did as well which might be of note.

Precision alignment of the LIGO 4 km arms using dual-frequency differential GPS

LIGO, however, posed several unique challenges. The beam tubes needed to be aligned along the propagation direction of light in vacuum and not along the direction perpendicular to local gravity on the surface of the Earth11. The curvature of the Earth will cause the Earth's surface to deviate from the straight line propagated by light in vacuum by 1.25 meters over a 4 km path if the line starts out level with the surface...

The fundamental coordinate system for the alignment was the Earth ellipsoidal model WGS-8414,15. All raw GPS data were referred to this system using geodetic coordinates {height above ellipsoid [h], latitude [f], longitude [l]}. Geodetic coordinates were transformed to the standard earth-fixed Cartesian system {XE, YE, ZE}, where zˆ E is aligned along the earth’s polar axis and xˆ E penetrates the ellipsoid at the intersection of the Greenwich Meridian with the Equator. yˆ E is perpendicular to both axes (refer to APPENDIX A)...

Using global coordinates, the beam tube centerlines were marked along the foundation slab at points spaced uniformly at ~20 m intervals (the unit length of beam tube sections that were welded together in the field) along both arms.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #166 on: February 14, 2022, 10:53:29 PM »
You might find this interesting regarding lasers over distance. Specifically how the 4KM long arms of the two LIGO sites were measured and constructed creating a straight light path in relation to Earth's curvature. There are some transformations they did as well which might be of note.
...
Thanks. You never quite stop to think about all the remarkable feats required for these projects.
In my earlier models, earth was just a cylinder and celestial objects were traveling circles. Initially i even had the sun going backwards
In my later ones, i am indeed using WGS84 data to more accurately calculate positions of celestial objects however my earth model is still a cylinder.
Converting the cylinder to WGS84 would create a little bulge around the equator-ring but the entire model would remain perfectly circular. (lat/long goes from [0-2π] regardless of obloidness)
However I'm currently not very keen on making this change as i suspect it will be computationally too expensive and it's probably not that visually interesting.
But thank you for this nugget of information

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #167 on: February 15, 2022, 12:00:42 AM »
Quote
If you insist on measuring reality with an orthogonal ruler and drawing it on an orthonormal axis, you will indeed get a sphere. But then you are pre-supposing an orthonormal axis.

Orthonormal coordinates are not the only coordinates in Euclidean space.  Just because the coordinate lines are curved, does not mean that the space is curved.  The coordinate system does not define the whether or not a space is curved.

If a space has intrinsic curvature, the Riemann curvature tensor is non-zero.  What is the value in your model?

The existence of non-zero curvature can be easily visualized by parallel transport.

A sphere has it because it is possible to change the direction of vector by forming a loop over the surface.  A flat disk does not have it because you can take any random path and when you return to the origin, the vector will have the same direction.

It’s an easy test and you don’t have to assume any shape or even take any measurements. Can you show a parallel transport on you model?

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #168 on: February 15, 2022, 03:44:05 AM »
Quote
1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original.
3. Flattening a sphere into a disk is a coordinate change.
4. Therefor, a disk is mathematically equivalent to a disk.
5. Because a disk is mathematically equivalent to a sphere, the physics is equivalent.
6. Because it is physically equivalent, light must bend however it needs to to change the RE appearance into FE reality.
7. Because the disk and sphere are mathematically equivalent, the equation for distance on a sphere can be used on a disk.
8. If the distance calculation on a disk does not match observed reality, then the process of measurement needs to be flexible to be made to match, measurement is not being done right

1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original. They are only equivalent if the change was isometric.
3. Flattening a sphere into a disk is a coordinate change. Flattening a sphere is more than just a coordinate change because it also changes the metric and the Riemann curvature tensor
4. Therefor, a disk is mathematically equivalent to a disk. When you transform a disc to a sphere the metric changes.  The Riemann tensor changes from 0 to non-zero.  Therefore, a disk and a sphere are not mathematically equivalent.

#### FLATEARTHPSYOP

• 4
##### Re: Found a fully working flat earth model?
« Reply #169 on: February 15, 2022, 06:41:51 PM »
Troolon ,  in Blender 3.0 --- would you be able to share me the source code in which I can project t he light paths of a sphere on different geometries that you have shown?
Very compelling

Any guides you can show me on how to do this this, just what you have done? Thanks for your hard wrok

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #170 on: February 15, 2022, 09:33:25 PM »
Troolon ,  in Blender 3.0 --- would you be able to share me the source code in which I can project t he light paths of a sphere on different geometries that you have shown?
Very compelling

Any guides you can show me on how to do this this, just what you have done? Thanks for your hard wrok
I've made all graphics with python and vpython (https://www.vpython.org/) (a webgl based lib).
However i've seen blender also supports python so maybe, something is possible?
Blender seemed to have a rather steep learning curve though which is why i started with vpython.
A bit more details can be found on https://troolon.com, but if you would like bits of code or practical advice on how to do this, perhaps shoot me an email? troolon _at_ the website i just mentioned: (troolon.com)

BTW here's my latest graphic: a view of the solar systems showing planet positions from jan 2022 up to dec 2022.

#### WTF_Seriously

• 1342
• Nobody Important
##### Re: Found a fully working flat earth model?
« Reply #171 on: February 15, 2022, 10:03:37 PM »
BTW here's my latest graphic: a view of the solar systems showing planet positions from jan 2022 up to dec 2022.

Looking at this, how is it that the distance from Sydney to LAX is a little over 12,000 km yet the distance from Sydney to Santiago is a little over 11,000 km.
I hope you understand we're maintaining a valuable resource here....

#### DuncanDoenitz

• 444
##### Re: Found a fully working flat earth model?
« Reply #172 on: February 15, 2022, 10:22:00 PM »
You're using a straight ruler.

Fully working model, but you need a fully working stretchy-bendy ruler.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #173 on: February 15, 2022, 10:33:55 PM »
Orthonormal coordinates are not the only coordinates in Euclidean space.  Just because the coordinate lines are curved, does not mean that the space is curved.  The coordinate system does not define the whether or not a space is curved.

If a space has intrinsic curvature, the Riemann curvature tensor is non-zero.  What is the value in your model?

The existence of non-zero curvature can be easily visualized by parallel transport.

A sphere has it because it is possible to change the direction of vector by forming a loop over the surface.  A flat disk does not have it because you can take any random path and when you return to the origin, the vector will have the same direction.

It’s an easy test and you don’t have to assume any shape or even take any measurements. Can you show a parallel transport on you model?
I think it's explained here: starting at 10:24 till 22:28
Rieman tensor: Rθφθφ = r²sin²φ
Ricci scalar R = 2/r²   (r = 6000km)

But to understand this more intuitively i must go again to to the construction. ...
2. We convert the sphere to geographic coords (lat, lon, dist). The graphic looks the same, angles and distances still measure the same. It still looks like this:
3. draw latitude on a straight axis instead of a radial one. This is just a change in representation. We do not change coordinates, distances or angles. There's no change in intrinsic curvature as this step is invisible to the maths. I've attached a graphic of what i think the parallel transport looks like but i'm not 100% certain.
Ultimately this graphic is irrelevant as this last step is just a change in representation (like pie charts vs bar charts and not in coordinates, distances or angles -- at least not intrinsically).

Now if you answer, please tell me which of these two steps is incorrect rather than having me learn up on another field of complicated math
Physics switches between cartesian and celestial coordinates all the time. If this breaks angles and distances would break here, a large part of physics would be broken.
« Last Edit: February 15, 2022, 10:52:29 PM by troolon »

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #174 on: February 15, 2022, 10:38:50 PM »
Looking at this, how is it that the distance from Sydney to LAX is a little over 12,000 km yet the distance from Sydney to Santiago is a little over 11,000 km.
This thing is not in cartesian coordinates (x,y) but in celestial coordinates (latitude,longitude,distance).
When you calculate the distance between places using lat/lon coordinates, you will get the same distances as on a globe. (think haversine-formula)

This graph is not in an orthonormal basis, you can't take a ruler to it and just measure it. Mathematically it checks out though.

#### Dr Van Nostrand

• 1234
• There may be something to this 'Matrix' stuff...
##### Re: Found a fully working flat earth model?
« Reply #175 on: February 15, 2022, 10:59:44 PM »
Hey,

This is how you make the any shaped earth:
- coordinate transformations can change the universe to any shape you like
- physics is unaffected by coordinate transformations
-> physics can be made to work on any shape world (see the animations i posted for example)
-> There's no test to distinguish the models, they're designed to be identical
-> It's impossible to ever know the shape of the planet

I drove a car all over and across North America for business travel for many years. I had to track my mileage and gas usage for expense accounts. I know exactly how big North America is.

There are people in Australia who have done the same thing. They know how big their continent is.

Distances can be measured.

What you are really saying is that it is impossible for YOU to know the shape of the Earth. I'm guessing you just don't travel a lot.
Round Earther patiently looking for a better deal...

If the world is flat, it means that I have been deceived by a global, multi-generational conspiracy spending trillions of dollars over hundreds of years.
If the world is round, it means that you’re just an idiot who believes stupid crap on the internet.

#### WTF_Seriously

• 1342
• Nobody Important
##### Re: Found a fully working flat earth model?
« Reply #176 on: February 15, 2022, 11:21:32 PM »
Looking at this, how is it that the distance from Sydney to LAX is a little over 12,000 km yet the distance from Sydney to Santiago is a little over 11,000 km.
This thing is not in cartesian coordinates (x,y) but in celestial coordinates (latitude,longitude,distance).
When you calculate the distance between places using lat/lon coordinates, you will get the same distances as on a globe. (think haversine-formula)

This graph is not in an orthonormal basis, you can't take a ruler to it and just measure it. Mathematically it checks out though.

I'm not calculating anything.  I'm looking at your picture.  When you look at the disc, straight line path from Syndney to Santiago goes right past LA and continues for a significant distance.  It's a simple observation.  No math involved.
I hope you understand we're maintaining a valuable resource here....

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #177 on: February 15, 2022, 11:25:16 PM »
Distances can be measured.
What you are really saying is that it is impossible for YOU to know the shape of the Earth. I'm guessing you just don't travel a lot.
Problem is my proof is mathematical by nature, so unfortunately  i've proven it's impossible for you to know the shape of the earth too
Physics is a model of reality. It's just a representation. You can have different representations of the same reality.
All i've done is constructed a different representation of physics...
So when you drive your car and measure distances that way, both representations will predict the same answer.
Soo.. now that we have different shaped representations of reality that can't be distinguished by any measurement, how will you tell what shape planet you're truly on?

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #178 on: February 16, 2022, 12:55:30 AM »
Quote
But to understand this more intuitively i must go again to to the construction. ...
2. We convert the sphere to geographic coords (lat, lon, dist). The graphic looks the same, angles and distances still measure the same. It still looks like this:
3. draw latitude on a straight axis instead of a radial one. This is just a change in representation. We do not change coordinates, distances or angles. There's no change in intrinsic curvature as this step is invisible to the maths. I've attached a graphic of what i think the parallel transport looks like but i'm not 100% certain.
Ultimately this graphic is irrelevant as this last step is just a change in representation (like pie charts vs bar charts and not in coordinates, distances or angles -- at least not intrinsically).

You are making this much more difficult than it is.  Intrinsic curvature is coordinate independent.  A sphere has it, regardless of what coordinate system you use..

Quote
What is meant is the intrinsic curvature of the space, meaning it is independent of the choice of coordinates. There are clever methods of determining whether and to what extend your space deviates from flat euclidean space, namely Gaussian curvature, and, more importantly, the Riemann tensor.

https://physics.stackexchange.com/questions/290906/what-is-really-curved-spacetime-or-simply-the-coordinate-lines#:~:text=Coordinates%20are%20most%20definitely%20curved,to%20zero%20in%20flat%20spacetime.

You are correct that transforming from Cartesian to Geo graphic coordinates doesn’t change the intrinsic curvature, but just drawing latitude on a straight axis doesn’t change it either.

The problem comes when you try and “flatten” the sphere...without any distortion.  It’s been pointed out to you umpteen times.  I’m not sure why you are having difficulty grasping it.  It is a fundamental principle of geometry that has been known for 100s of years that it is impossible to accurately represent a sphere on flat plane.  Trying to do that isn’t just a coordinate change, it is changing the fundamental properties of the geometry of the sphere.  Simple coordinate changes don’t do that.  As you say, a coordinate change is just another way of representing something.  If you try and flatten a sphere you aren’t changing the way something is represented, you are trying to represent something else

A flat disk has no intrinsic curvature.  The Riemann tensor will be zero.  A sphere has a non-zero Riemann tensor and it has intrinsic curvature.  A body can’t be a disk and a sphere at the same time.  It can only be one or the other and it is easy to determine which simply by doing a parallel transport.

On the left, the vectors make the full loop around the triangle, in the same direction, while not deviating from parallel.  On the right, the vector has to change direction in order to stay parallel all the way around the triangle.  If you look at it, its easy to see why. To stay on the surface, the vector has to “bend” around the top of the triangle.  No bending required on a flat triangle.  No complicated math, or measurements involved.

If there is no path on your model where a vector has to change direction in order to make a closed loop and staying parallel, then your model is flat.  It has no intrinsic curvature and is not mathematically, or any other way, equivalent to a sphere. You certainly have the graphic skills, so it shouldn't be hard for you to post evidence of doing a parallel transport.

Quote
Problem is my proof is mathematical by nature, so unfortunately  i've proven it's impossible for you to know the shape of the earth too

You keep contradicting yourself by stating that your model has intrinsic curvature and also stating that we can’t know the shape of the earth.  The very definition of intrinsic curvature is that “the inhabitants” can know.  So which is it?  Does your model have intrinsic curvature or is it impossible to know the shape of the earth?

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #179 on: February 16, 2022, 01:58:37 AM »
Troolon,

I would like to know which physicists agree with your ideas. I doubt any would say the earth could be any shape, or that changing coordinate systems changes shapes yet they are still equivalent, or that measurement varies.

You do not understand the concept of coordinate systems. I can't fix this, because 1. no time to take you through geometry, trig, solid geometry, and linear equations, and 2. you don't want to know a truth that would invalidate your theories. I don't know who the experts were, or what you said to them, what they said back yo you, or that you correctly understood their answer, but your ideas on coordinates, measurement, and physics are not true. This is fine on TFES, but try to take these ideas to the real world of physics and math, real experts, and they will say the same as I and others on this thread. If you convince them your ideas are true, you will be hailed as the greatest mathematician/physicist of all time. You are wrong, because you have to be to allow FE, which is why we know the earth is round.

You are doing the equivalent of looking at a funhouse mirror and seeing your legs looking a foot long and deciding you just don't know how long your legs are, any mirror could be right. Graph a ruler on cartessian, then logarithmic. The graph on cartessin will be isometric, linear, a multiple of the straight line. The graph on some coordinate system might be curved, but the ruler is still straight. It is not "could be any shape, no one knows". Please enroll in geometry class.

Gps satellites work by broadcasting a timestamp on each transmission. Your gps device has a table of where the satellite is and a very accurate clock. Subtracting the timestamp in the transmission from the time in the device gives the elapsed time, multiply by speed of light, and you get distance from a known location. This is a sphere around the satellite. Do this with four satellites, and you can calculate 4 spheres, which will intersect at only one point.

This depends on radio waves going straight and the speed of light being constant. If this is true Under your theories, we can't be sure of either. gps would be impossible. Yet it works amazingly well.

Do radio waves travel straight and is the speed of light constant? If not, how does gps work?

If I buy a lidar measuring gadget at Home Depot and take it to Australia, will it still work correctly? I think it will.

Bear in mind that you can buy a usb gps receiver and download open source gps software. You can examine the algorithms and look at raw data. There are web sites where you can look at the current locations of gps satellites and see their transmissions. If you know where satellite is and you can map the locations on the surface of the earth, the result will be a sphere.

The question boils down to: Australia too big on FE, just right on RE. Is the earth round, or is measurement impossible or somehow variable in ways that no one noticed, detectable only by the observing that straight light doesn't work on FE and completely unexplained?

Do you acknowledge that gps and lkidar devices work and match RE theory, while FE is not consistent with observed results without "fudge factors", as the FAQ says, "unknown forces with unknown equations"?

Where is Sigma Octantus?

a. no one knows
b it is everywhere (southern hemisphere) and nowhere (northern hemisphere)?
c ?
d. 204 light years in the direction of the south pole of round earth

Still waiting for your graphics to show sunset in Denver and how Salt Lake City sees daylight over the entire dome while St Louis sees night sky over the entire dome. Please show how someone at night looks up and see stars over the entire dome, including where the sun is. See right through the sun to the stars (beyond?) without seeing the light of the sun. Can you make a model that shows where sun and stars are, but from the point of view of someone on the surface?

Like this: https://stellarium-web.org/

But show how everything works when you move around the simulation on the surface. There's your homework. Explain gps with variable light speed and curved rays, and incorporate day/night sky as seen from the surface into your 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.