#### AATW

• 6640
##### Re: Found a fully working flat earth model?
« Reply #60 on: February 02, 2022, 12:24:53 PM »
And what does reality really look like? We don't know.
Yes we do. It has been measured and observed and there's only one shape which makes sense with those observations and measurements.
All you've done is show that you can map the reality on to different shapes.
People have been doing that for centuries with different map projections.
This is a good site which lets you play around with one such projection and see how the size of countries varies depending on where you place them.

https://thetruesize.com/

For example, Greenland isn't really bigger than Australia. It's just further north than Australia is south and thus more distorted by the projection.

Quote
Ultimately a model is just a representation of reality and you can represent it in infinitely many ways.
Sure. But only one model makes sense without distortion. If you map known distances, land mass shapes and sizes and flight routes on to a sphere they make sense.
Try and do it on a flat plane and there is distortion. You seem to be getting round that by distorting the coordinate system itself.
I'd suggest an earth where a meter in Greenland is the same as a meter in Australia is far more likely.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #61 on: February 02, 2022, 12:28:02 PM »
Quote
Philosophical implications:
The trick above generalizes to all coord transforms. You can practically turn any shape into any other shape.
And physics, well they use coord transforms all the time.
So all of physics basically works on any shape universe

You are over simplifying the concept of general covariance.  A sphere has intrinsic Gaussian curvature, which by definition means it can be measured from “the inside” and is coordinate independent. It doesn’t vanish when you change coordinate systems.

Quote
A curvature such as Gaussian curvature which is detectable to the "inhabitants" of a surface and not just outside observers. An extrinsic curvature, on the other hand, is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides.

https://mathworld.wolfram.com/IntrinsicCurvature.html#:~:text=A%20curvature%20such%20as%20Gaussian,surface%20on%20which%20he%20resides.
Quote
Take the curvature of the slice-curve. Thus you get a curvature in every direction. The largest (positive, upward) and smallest (most negative) such curvatures are called the principal curvatures, and they occur in orthogonal directions. Their average is called the mean curvature, and their product is called the Gauss curvature. For the unit sphere, both principal curvatures are 1 and hence the Gauss curvature is 1. For a unit cylinder, the principal curvatures are 1 and 0 and hence the Gauss curvature is 0. For a suitable hyperbolic paraboloid as in Figure 2, the principal curvatures are 1 and −1, and the Gauss curvature is −1.

https://www.ams.org/publications/journals/notices/201602/rnoti-p144.pdf

A flat sheet of paper has zero Gaussian curvature. If you take that sheet and roll it into a cylinder its Gaussian curvature stays zero.
Now take that sheet and wrap it over a sphere. You have to crease or tear the sheet to fit it around the sphere. That's because since a sphere has positive Gaussian curvature the circumference of a circle drawn on a sphere is less than 2 pi times the radius. You have to wrinkle the paper to get rid of the extra circumference.

One easy way to test for curvature on the earth is for two airplanes to start out some distance apart at the equator flying due north on different lines of longitude.  As they progress, their east-west separation decreases and eventually they meet at the north pole. That wouldn’t happen on a flat plane. You can also test using parallel transport.

You are also completely ignoring the effect of any spacetime curvature.  Where there is acceleration, there is spacetime curvature, where there is spacetime curvature, there is gravity and where there is gravity, there is a round earth.

Spacetime curvature is also coordinate independent.  It doesn’t vanish when you change coordinate systems.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #62 on: February 02, 2022, 12:40:22 PM »
In your above image at the left, the Sun's rays project light in only a "downward" direction from a single point.

Would not a spherical Sun emit rays in all directions though? If this be the case, than wouldn't you have rays projecting sideways  from the Sun and then bending downwards (due to bendy light) towards the flat Earth surface that you show as not lit?

Really sorry, I did indeed miss your question.
Yes you are right, the sun emits rays in all directions. I mainly wanted to show day/night on earth, and only drew rays to dusk/dawn areas. (to not clutter the picture)
Also bending of light is only strong around earth, so you wouldn't see that much bending around the sun.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #63 on: February 02, 2022, 12:52:24 PM »
Yes we do. It has been measured and observed and there's only one shape which makes sense with those observations and measurements.
Only if you're presupposing an orthonormal basis. But orthonormal basis is an assumption and not a law of nature.
All you've done is show that you can map the reality on to different shapes.
Yes! And therefore we can't know the true shape.

Sure. But only one model makes sense without distortion. If you map known distances, land mass shapes and sizes and flight routes on to a sphere they make sense.
Try and do it on a flat plane and there is distortion. You seem to be getting round that by distorting the coordinate system itself.
I'd suggest an earth where a meter in Greenland is the same as a meter in Australia is far more likely.
You're again presupposing an orthonormal coordinate system. Mathematically the measurements make exactly as much sense in any of the other coordinate systems. You can't take an orthonormal ruler to a non-orthonormal base. If i take my curvy earth ruler to your sphere it also wouldn't make sense at all.

#### AATW

• 6640
##### Re: Found a fully working flat earth model?
« Reply #64 on: February 02, 2022, 01:59:51 PM »
Mathematically the measurements make exactly as much sense in any of the other coordinate systems.
But in physical reality, not so much.
What you have got is an interesting thought experiment or philosophical thought, nothing more.
I'm assuming that a meter in Greenland is the same as a meter in the UK is the same as a meter in Australia, and that a 90 degree angle is the same everywhere too.
I believe that's a reasonable assumption - that the scale of reality doesn't randomly change depending on where you are.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #65 on: February 02, 2022, 02:25:42 PM »
Quote
A curvature such as Gaussian curvature which is detectable to the "inhabitants" of a surface and not just outside observers. An extrinsic curvature, on the other hand, is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides.

You may have me beat there, though i can't say for sure. The math is way to advanced for me and i don't really understand the parallel planes example. In both the flat and sphere world they would end up at the northpole in my mind.

Could you perhaps help me understand at what point the math breaks?
- The first step is that i express everything in celestial coordinates(lat, long, distance). But I don't think this breaks curvature because earth is still a sphere and all distances are the same.
- The second step is that i draw latitude on a straight axis instead of an angular one. So numerically everything stays the same. How can the math then notice something changed? It can't see the drawing and the numbers didn't change?
« Last Edit: February 02, 2022, 02:27:34 PM by troolon »

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #66 on: February 02, 2022, 04:15:03 PM »
Quote
Quote
A curvature such as Gaussian curvature which is detectable to the "inhabitants" of a surface and not just outside observers. An extrinsic curvature, on the other hand, is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides.

You may have me beat there, though i can't say for sure. The math is way to advanced for me and i don't really understand the parallel planes example. In both the flat and sphere world they would end up at the northpole in my mind.

Could you perhaps help me understand at what point the math breaks?
- The first step is that i express everything in celestial coordinates(lat, long, distance). But I don't think this breaks curvature because earth is still a sphere and all distances are the same.
- The second step is that i draw latitude on a straight axis instead of an angular one. So numerically everything stays the same. How can the math then notice something changed? It can't see the drawing and the numbers didn't change?
Quick question, what does intrinsic curvature do with teleporters? When you walk of the edge in antarctica, you appear again at the other end of the map. Would such a teleport contribute to curvature?
When i do the bug-walking-a-circle-test as described in the paper, the tiny "geodesic" circle still behaves as on a globe. ie it's circumference is smaller than 2*π*R. (for an observer inside the universe)
I'm really really out of my debt in this field of maths so please excuse if me if i'm horribly misinterpreting everything. However i think the teleport at antarctica might possibly be hiding some curve.

I currently have two conflicting sentiments:
- Your claims about intrinsic curvature which i can't even pretend to be knowledgeable enough to assess let alone start speculating about the effect of teleporters on the math.
- The fact that there are no observable differences between the two models. You said intuitively the differences in distances and angles give the shape away, but there are no observable differences with the globe model. All distances and angles measured are the same in both models (for an inside observer)

#### Iceman

• 1825
• where there's smoke there's wires
##### Re: Found a fully working flat earth model?
« Reply #67 on: February 02, 2022, 04:23:12 PM »
Can you elaborate on this teleporting across the ‘edge’ of Antarctica in relation to curvature? What exactly do you mean?

#### WTF_Seriously

• 1342
• Nobody Important
##### Re: Found a fully working flat earth model?
« Reply #68 on: February 02, 2022, 04:47:23 PM »
Imagine i have 2D cartesian coordinates (0,0) and (4, 4).  Then the distance formula would be sqrt((x1-x2)² + (y1-y2)²
In polar coordinates these coordinates would be (0°, 0) en (45°, 4).  If i plug these numbers into the distance formula, i get total gibberish. When you do a coord transform, you must update all formulas. That is what i mean with a new distance metric. It's the old one with compensation for the coord transform.

First off, the polar coords would be (0°, 0) and (45°, sqrt(32)).  Your radius from 0 is not 4.

As folks have been trying to tell you, distance is distance.  Changing the coordinate system doesn't change that.  It just changes how you are describing it as well as how you'd calculate it but it doesn't change what it actually is.
I hope you understand we're maintaining a valuable resource here....

#### Tumeni

• 3179
##### Re: Found a fully working flat earth model?
« Reply #69 on: February 02, 2022, 05:07:21 PM »
Consider this thought experiment: that shows the earth can have any possible shape ...  We will simply never know the shape of the planet.

Consider this thought experiment;

In the late 1950s, Russia launched the first orbital satellite; Sputnik 1. This was observed by scientists and public the world over, and its radio transmission, coming and going, confirmed this was not merely a natural phenomenon. It was observed, and heard, to pass every 90 minutes or so.

A number of years pass, other space flights take place, and in the present day, the ISS passes over us every 90 mins. We can watch it go by, we can time the interval between two passes in one night, and find that this is around 90 minutes, same as Sputnik.

Hundreds of humans have completed orbital flights, and have spent time on the ISS. A select few have gone to the Moon and back.

Does your model account for the anecdotal, photographic, geological and other proofs of all these missions?

Basically; if the Earth was NOT a globe, dontcha think someone out of these hundreds of folk would have noticed ....?
=============================
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 #70 on: February 02, 2022, 05:40:27 PM »
What you have got is an interesting thought experiment or philosophical thought, nothing more.
agreed. The model will not be of much use in practice. Just the philosophical conclusion that earth can have any shape and we'll never know was rather surprising to me when i stumbled onto it.
I'm assuming that a meter in Greenland is the same as a meter in the UK is the same as a meter in Australia, and that a 90 degree angle is the same everywhere too.
I believe that's a reasonable assumption - that the scale of reality doesn't randomly change depending on where you are.
I want to clarify that for an observer inside the flat model, all these measurements will match up.
But for you as an outside observer that's indeed a very handy property
One small caveat: when i'm driving to the supermarket, the flat model does match my intuition more.
My interpretation of  occams razor is: If you have 2 models, take the one that makes your job easiest, and you're quite correct, that will almost always be the globe model.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #71 on: February 02, 2022, 06:03:12 PM »
Can you elaborate on this teleporting across the ‘edge’ of Antarctica in relation to curvature? What exactly do you mean?
As you know we're mapping the globe model onto a flat earth. On the globe a hiker can happily walk across the southpole.
Plotting this trajectory on the AE-map the hiker will appear to suddenly teleport to other side of the disc.
Transforming with any other point than the northpole on top, will make antarctica continuous again, so this is just a mathematical artifact.
But for the effect of curvature, it does mean that the greatcircle northpole-southpole-northpole  is really a closed loop and not and open line.
- distances and angles in the model all match the globe model for an inside observer. So i'm not sure where difference in curvature would come from
- Doing the bug-walking-a-circle and comparing with 2*π*R also seems to confirm it curves like a globe,
- by construction i can't readily identify when the curve-breaking would be happening.
- when we shift the axis slightly away from the southpole, it will suddenly become a close curve again. So i think this is asymptotic behaviour of a loop and i think the intrinsic curvature is still 1/R² in the limit.
- in general, math doesn't like teleporting

Again i'm in way too deep over my head, so i can't draw any conclusions. Just trying to match it with my intuition.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #72 on: February 02, 2022, 06:10:43 PM »
As folks have been trying to tell you, distance is distance.  Changing the coordinate system doesn't change that.  It just changes how you are describing it as well as how you'd calculate it but it doesn't change what it actually is.
What do you mean with distance?
I suppose we're comparing distance measured between 2 points in reality with:
- distance measured on the outside of the model with an orthogonal ruler
- distance measured on the outside of the model with a flat-earth ruler
- distance measured by someone living inside the model
- mathematically calculated distance.
Only the first method fails. All 3 others give the correct result.
I've already conceded that it's indeed a handy property for a model to be measurable with an orthogonal ruler.
However the entire purpose of the model is philosophical in nature and for that we only want to show that "it works", which it does.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #73 on: February 02, 2022, 06:17:23 PM »
Does your model account for the anecdotal, photographic, geological and other proofs of all these missions?

Basically; if the Earth was NOT a globe, dontcha think someone out of these hundreds of folk would have noticed ....?
The model i present is the globe model transformed.
It turns out physics works regardless of shape (it's math, it only cares about the numbers) and it can be proven there is no test that can differentiate between the 2 models.
So if the globe model can account for all the evidence, so can mine.

#### jimster

• 297
##### Re: Found a fully working flat earth model?
« Reply #74 on: February 02, 2022, 09:09:11 PM »
Physics is looking at the world around us and developing a consistent explanation of what we see. Math is starting with some assumption and developing a logically consistent system. Physics is, math describes. The same physical object can be described in either cartessian or polar coordinates, but a sphere is still a sphere and a plane is still a plane. The coordinates are not bounded, 3 space to infinity is within both. The reeason locations on the globe are given as latitude and longitude, i.e. polar coordinates, is that the earth is round. If it were flat, we would use x/y coordinates, much easier.

If you transform a sphere into a plane or vice versa, points on the plane will be different distances, per Gauss. If you can make a flat map with constant scale, the earth is flat. If you can make a spherical map with accurate distance, direction, and scale, the earth is not flat, or Gauss' theorem is not true. There is no flexible measuring in geometry, the ruler is straight and constant. If you need to bend or stretch the ruler, you are proving Gauss' point.

When you flatten the globe into a disk as in the FAQ map, mathematically each latitude gets longer all the way to the south pole. Do car odometers in Australia measure distance differently from those in EU? Planes fly faster and have longer range? Do rulers stretch as they travel south?

Bending the light is, as the EA page in the FAQ says, "unknown forces with unknown equations". Making the ruler curved and the scale adjustable by location is fudge factor without any justification, what psychologists call "motivated reasoning". You get there by observing that assuming FE produces bad results, so you hypothesize fudge factor without proof in order to save your belief.

Your model shows how we could see sunset/sunrise and day/night on FE, assuming some directional phenomenon and the unexplained bending of light, coincidentally exactly the equations to transform RE into FE. But you're not done with explaining all phenomena we observe, we all see the dome, geometrically we all see all of it. Yet at the same moment, some see stars all over the dome, others see light blue all over the dome, they can be as close as perhaps 300 miles. Someone in the northern Hemisphere sees completely different stars than southern hemisphere.

Then there is Sigma Octatis, the southern (pretty close) pole star. At the summer (northern hemisphere) equinox, at midnight in South Africa it is just after sunset in western Australia and just before sunrise in South America. You can see Sigma Octatis directly south from all these places at that time. On FE disk map, Sigma Octatis is in directly opposite directions from South America and Australia.

So please show the dome appearance in your model. I will be interested to see: Where is Sigma Octatis and how do the light rays from it travel? How does the dome appear light blue for some, and for those who see it as dark, different stars. How do those stars appear to travel across the dome in different directions at the same time? Please show with your model.

Do you agree that flattening a sphere into a disk will geometrically distort the distances? Or is Gauss' theorem not true?

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.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #75 on: February 02, 2022, 11:37:12 PM »
If you transform a sphere into a plane or vice versa, points on the plane will be different distances, per Gauss. If you can make a flat map with constant scale, the earth is flat. If you can make a spherical map with accurate distance, direction, and scale, the earth is not flat, or Gauss' theorem is not true. There is no flexible measuring in geometry, the ruler is straight and constant. If you need to bend or stretch the ruler, you are proving Gauss' point.
When you flatten the globe into a disk as in the FAQ map, mathematically each latitude gets longer all the way to the south pole. Do car odometers in Australia measure distance differently from those in EU? Planes fly faster and have longer range? Do rulers stretch as they travel south?
I believe everything works because of the distance metric changing. My guess is that the changes to the distance metric nullify the changes in shape and i believe i can make a compelling case for this.
- The distance metric on a globe is haversine. This is a function that takes lat/long as input. lat/long are invariant on the AE map, and so the AE map with haversine as a distance metric will return the same distance between any 2 points on the map as a globe does. The formula to calculate angles on a globe is also expressed in function of lat/long coordinates and is again invariant under AE. So in these 2 maps, all distances and all angles are the same as on the globe.
As gauss measures curvature based on distances, i really can't see how the curvature can change.
- I thought about simulating distances in my little simulation, and didn't because it's pointless. I would be doing haversine in both cases on the same numbers...
- The bug-looping-a-circle-method to find curvature (if it's less than 2*π*R, also gives the same result in my world)
- and finally a proof by construction: construction of the flat world happens in 2 phases: transform to celestial coords(lat, long, dist) and then draw the latitude on a linear axis instead of a radial one (like AE). I do not believe the drawing on a different axis can cause curvature to break. This drawing is just a rendering of the results. The coordinates don't change. Math can't see how i draw things.
However i believe the switch to celestial coordinates might be the problem. When we switch from cartesian to celestial coordinates, all coordinates change, and the distance formula breaks. That is why we switch the distance to haversine at this point. If we draw a sphere in cartesian coordinates, and then express it in celestial coordinates, while changing the distance metric, the sphere is still a sphere. If we only changed the coordinates without changing the distance function, gaussian curvature would probably break.
So i believe gaussian curvature is dependent on shape and distance. If we change the shape and compensate with the distance, this works. Of course in reality this also works. People have been using haversine to calculate spherical distances for a very long time, and if the sphere would cease to be when you transform from cartesian to celestial, that would be rather problematic.

Quote
Bending the light is, as the EA page in the FAQ says, "unknown forces with unknown equations". Making the ruler curved and the scale adjustable by location is fudge factor without any justification, what psychologists call "motivated reasoning". You get there by observing that assuming FE produces bad results, so you hypothesize fudge factor without proof in order to save your belief.
Why does light travel straight on the globe? (Because it matches observations?) Same with bendy light. Again, it's the globe model transformed. The explanation is the same. As for "motivated reasoning", can you give me the reason why only an orthonormal basis may be considered to describe the universe?

Quote
Your model shows how we could see sunset/sunrise and day/night on FE, assuming some directional phenomenon and the unexplained bending of light, coincidentally exactly the equations to transform RE into FE. But you're not done with explaining all phenomena we observe, we all see the dome, geometrically we all see all of it. Yet at the same moment, some see stars all over the dome, others see light blue all over the dome, they can be as close as perhaps 300 miles. Someone in the northern Hemisphere sees completely different stars than southern hemisphere.
It works the same way as in the globe model, but then transformed. If you'd like to get a visual, draw the scene on the globe, draw light-rays, transform. If the globe can explain it, so can we.

Quote
Then there is Sigma Octatis, the southern (pretty close) pole star. At the summer (northern hemisphere) equinox, at midnight in South Africa it is just after sunset in western Australia and just before sunrise in South America. You can see Sigma Octatis directly south from all these places at that time. On FE disk map, Sigma Octatis is in directly opposite directions from South America and Australia.

So please show the dome appearance in your model. I will be interested to see: Where is Sigma Octatis and how do the light rays from it travel? How does the dome appear light blue for some, and for those who see it as dark, different stars. How do those stars appear to travel across the dome in different directions at the same time? Please show with your model.
Same answer as above. If it works on a globe, it works on the flat earth as it is the globe model transformed. I will not be rendering out all examples you've requested. We believe this model is only of philosophical use and thus quite pointless to do so. But i've explained all formulas, feel free to investigate further. However i have rendered the south star before. Couple of notes on the picture: All points below the southpole are transformed into a circle (thank you AE map). So the yellow circle you see is the star. The colored lines are the sightlines from the 3 observers. It's easy, draw on a globe, add lightrays from source to destination, transform. The other pictures show from where the southstar is visible and the shared area of the sky the 3 observers can see (it's the area inside the yellow circle and outside the colored ones). I hope it makes sense

Quote
Do you agree that flattening a sphere into a disk will geometrically distort the distances? Or is Gauss' theorem not true?
I believe distances only become distorted depending on the distance metric. Haversine is valid for both models, but generally, all formulas need to be coord transformed (as one typically does in physics when doing coord transforms...)
« Last Edit: February 03, 2022, 12:09:29 AM by troolon »

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #76 on: February 02, 2022, 11:56:02 PM »
Quote
I believe everything works because of the distance metric changing. My guess is that the changes to the distance metric nullify the changes in shape and i believe i can make a compelling case for this.
Quote
I believe distances only become distorted depending on the distance metric. Haversine is valid for both models, but generally, all formulas need to be coord transformed (as one typically does in physics when doing coord transforms...)
I stand corrected. It's not only the distance metric that's undoing the curvature change. During the cartesian_to_celestial coord change: 3 things happen:
- The axis changes
- all coordinates change
- the distance metric changes
It's the combination of these 3 factors that makes sure the universe doesn't change curvature.
Take a sphere, go from (x,y,z) to (lat,long,dist) and it will still look like a sphere. So coord transforms, when done correctly, don't distort curvature. The rest of the reasoning in the post above was correct though.

Also i'd like to reiterate the goal here:
We've created a thought experiment that shows that:
- the shape of the universe can be changed, without affecting the physics
- it is impossible to differentiate between the different shaped models. There exists no test, observation or measurement to find a difference.
- The universe can have many different shapes and we'll never be able to prove which one. Flat and globe are both possible though. As is a simulation, or no shape or ...
And as a side-effect we've also created a fully working flat-earth model.
« Last Edit: February 03, 2022, 12:12:46 AM by troolon »

#### jimster

• 297
##### Re: Found a fully working flat earth model?
« Reply #77 on: February 03, 2022, 03:41:35 AM »
The haversine gives the correct distance over the surface of a sphere between two points. When you transform this onto a 2d circle as in the FAQ map, you can have each point on the same lat/long as on a sphere. But those points are no longer on a sphere where the lat/long are both polar coordinates. Latitude is now along a straight line, you can't do trig with a straight line. Lat/long no longer has the meaning that makes haversine valid. All you are ever doing is getting the same answer as you would on a sphere. But you are no longer on a sphere, haversine means nothing when it is not segments of a curved surface. Of course you get the same answer, all you did was plug the same numbers into the same formula, but the formula does not apply to the product of the transformation. The distance must be calculated as the distance between the endpoints of the arc.

It is as if you took a flat map and determined a distance  by d = sqrt( (x2 - x1)squared + (y2 - y1)squared ), map it onto a sphere, label each point with the x,y position and then use the same formula with the same numbers, you get the same answer. But it isn't right, because the formula is only valid on a plane.

Meanwhile, in the real world, if you stretch latex over a sphere and paint the land masses of the earth on it, you can make a map with correct distances (haversone works). If you peel it off from the south pole up and flatten it, the southern hemisphere will have to stretch out. This is why Australia is half again as wide on the FAQ map. Now it is flat, though, and you need (angle. distance) and use 2d polar coordinate trig. Then you will get a number that matches the appearance/physical measurement of your transformed flat map. No stretchy ruler needed, it is a different distance, per Gauss.

Once again, haversine works on a sphere. When you flatten it, you have to use polar coordinate trig. Distance is not maintained through your transform. Latitude on FE map is not degrees on a sphere, it is just the relative distance from the center.  The haversine formula needs the angle between both latitude and longitude, but there is no such angle for latitude on a flat map.

Again, using haversine on a flat surface does not produce a right answer. . This is the error of your logic re "distance is preserved". Gauss says it is not, so do the faq maps that clearly show Australia bigger and Greenland smaller on the flat map. So does your model.

Where on your model is Sigma Octatus? Can you show startrails valid from every point? Can you show the entire dome visible to everyone, some with stars and some with daylight and sun? The entire dome filled with stars, yet completely different stars northern vs southern hemisphere? RET can, that's why the earth can only be round.

Distance is not preserved through your transform, RET accounts for all these things. Any other shape needs the light to bend due to "unknown forces with unknown equations". That's why we know the earth is round.

Repeat with me, haversine only works on a 3d sphere, not a 2d disk. The correct formula produces the exact thing you see in the faq map and your model. No distance funny business needed.

Sorry if I repeated myself, it is the natural human tendency when people on this site can't or won't understand.
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.

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #78 on: February 03, 2022, 03:50:50 AM »
Quote
Quick question, what does intrinsic curvature do with teleporters? When you walk of the edge in antarctica, you appear again at the other end of the map. Would such a teleport contribute to curvature?

Basically, parallel transporting means that on a flat plane you can take a vector and send it on any arbitrary closed loop path, keeping it parallel with itself. When it returns to the origin, it is unchanged.  Not so with a sphere.  When it returns to the origin, it will have changed directions. It’s the “the way” to determine intrinsic curvature and is used the define the Riemann curvature tensor.

http://www1.kcn.ne.jp/~h-uchii/Einstein/riemann.curv.html
Here’s a demonstration by Brian Greene.  He kind of rambles in the beginning, but stick with it.

https://www.britannica.com/video/222305/Your-Daily-Equation-27-Curvature-and-Parallel-Motion

Here’s a more indepth  explanation that goes into the very complicated math

If the Riemann curvature tensor is zero in one coordinate system, it is zero in all of them.  If it is non-zero in one coordinate system it is non-zero in all of them. The video also goes into Geodesic deviation as a way to identify intrinsically curved space. You don’t have to watch much of it to realize that identifying  whether or not a space has intrinsic curvature is more complicated that just looking at it.

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When i do the bug-walking-a-circle-test as described in the paper, the tiny "geodesic" circle still behaves as on a globe. ie it's circumference is smaller than 2*π*R. (for an observer inside the universe)

Then that means whatever surface you were testing is intrinsically curved, but not knowing exactly what you did or how you did it...I can’t know for sure.

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I'm really really out of my debt in this field of maths so please excuse if me if i'm horribly misinterpreting everything. However i think the teleport at antarctica might possibly be hiding some curve.

I can’t help you too much on that end, most of it is over my head as well.  However, I do know that you can’t transform an intrinsically curved space onto a flat surface without some deformation.  That should be intuitive.  Have you ever tried to gift wrap a basketball? You can’t do it without mutilating the paper.

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The implication of this is that there is no coordinate transform which maps a sphere into a flat plane without deforming it. This is a trivial counterexample to your question — you cannot make a spacetime look flat just by changing the spatial coordinates you are measuring it in!
You can find a transform that makes the metric locally inertial (i.e. Freefalling coordinates) but this is not globally flat. This is all because a manifold is independent of the coordinates you use to measure it. You can't make a sphere any less round by using a different ruler, or opening it up — it is just round shape!
The same cannot be said for a cylinder, however — which can be opened up into a flat plane — this indicates an important distinction between intrinsic curvature (the sphere) and extrinsic curvature (the cylinder).
This leads to the slightly counterintuitive assertion that the surface of a cylinder is a flat shape — in terms of its geometry, it is indistinguishable from a flat plane. Thus, it is flat!

Jack Fraser
DPhil Theoretical Physics University of Oxford (2018 - 2022)

https://www.quora.com/General-relativity-says-spacetime-is-curved-but-can-I-choose-coordinates-where-it-looks-flat

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it is impossible to differentiate between the different shaped models. There exists no test, observation or measurement to find a difference.

No its not impossible.  Just perform a parallel transport.  You might have to get creative with the logistics, but its doable.  In fact, a very similar concept was used to measure the curvature of spacetime using a gyroscope.

http://www.thephysicsmill.com/2015/12/27/measuring-the-curvature-of-spacetime-with-the-geodetic-effect/

Notice the gyroscope can't make the full loop without changing directions.

« Last Edit: February 03, 2022, 03:54:01 AM by Rog »

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #79 on: February 03, 2022, 03:58:08 AM »
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Can you elaborate on this teleporting across the ‘edge’ of Antarctica in relation to curvature? What exactly do you mean?

If that is directed at me, I think you misunderstand what "parallel transport" means.  See my response to Troolon above.