*

Offline stack

  • *
  • Posts: 3285
    • View Profile
Re: Found a fully working flat earth model?
« Reply #200 on: February 17, 2022, 05:51:31 AM »
Quote
Straight is only defined the way you think it is in an orthonormal basis.

Straight is the shortest distance between two points, regardless of the coordinate system.  Straight is a curve on a sphere.  What would be the shortest distance between NY and Moscow on your model?

If I understand the model, and Troolon, correct me if I'm wrong, I think it's pretty simple. It's a great circle, identical in both the RE and Troolon models. Same route, same distance. Just depicted, portrayed, visually different. Just like how the route would show as curved on a Mercator projection though it appears "straight" on a globe.

Offline Action80

  • *
  • Posts: 2006
    • View Profile
Re: Found a fully working flat earth model?
« Reply #201 on: February 17, 2022, 11:22:06 AM »
Being "inside" a coordinate system is a meaningless concept.
Why are coordinate systems used for travel?

Most county road numbering systems in the US specifically utilize plane geometry, commencing from county center and measured outward in .25 mile increments. North/South roads are oriented East/West and vice - versa.
« Last Edit: February 17, 2022, 12:12:25 PM by Action80 »
It's so hard to have faith in humanity when they do shit like this.

"I hate the police so I'm gonna burn a Walgreen's!"

Re: Found a fully working flat earth model?
« Reply #202 on: February 17, 2022, 01:35:06 PM »
The really impressive thing is how patient OP has been in explaining it in 10 pages of posts.

He had me worried for a minute talking about teleporting, but it was just a lack of a better word to describe some of the artefacts in looking at things after the coordinate transformation in the model

At least from what I've seen, most of the pushback is coming from the fact that OP claimed to have found a "working flat earth model". And that it's impossible to know the shape of the earth, it could a disk, globe, cuboid, Klein bottle, etc. What he/she did was a simple coordinate transformation, and therefore the representation is still of a globe. The pictures he/she showed are not representing a flat surface, but a sphere. Then everything else is made to fit that particular representation

Quote from: troolon
A disc in an orthonormal basis, with orthonormal distances, will have a curvature of 0.
A disc in celestial coords, with spherical distances, will have the curvature of a sphere. (as shown)
Also, this is wrong. As it has been mentioned, curvature is invariant over a change of coordinates. If it's non-zero, there is no possible change of coordinates that can turn it to zero globally.

Quote from: troolon
The model has intrinsic curvature but noone says an orthonormal axis is the only way it can be viewed.
Doesn't matter the axis or coordinate basis. If your model has non-zero intrinsic curvature, it isn't flat

*

Offline JSS

  • *
  • Posts: 1618
  • Math is math!
    • View Profile
Re: Found a fully working flat earth model?
« Reply #203 on: February 17, 2022, 02:20:29 PM »
I don't think that particularly matters, frankly. Aside from the entertainment value that Pete already mentioned, it may very well not have many practical applications (hence, the phrase "logically sound but practically useless" or however it was phrased earlier). But lots of things start out that way, especially in mathematics. String theory was considered useless to the physics world shortly after its creation, and it was only after many years that people started revisiting it and finding it might possibly have some practical applications.

Actually String Theory is a good example of my point earlier.  It actually has no practical applications, and can't have any predictive models based on it due to the impossibility of testing the accuracy.

The core issue with String Theory is that it states that the laws of the universe are based on how 10 (or 11) dimensions are folded up. The problem is that to use String Theory to predict the behavior of our universe we need to know how those dimensions are folded.

There are 10^200000 possible combinations and String Theory as of yet has no way to even begin to try and identify which one we live in.  It's like having a book that describes all the laws of physics, but it's encrypted and the key is the full text of the book. We can't prove String Theory is correct until we already know how everything works, and can test it all.

Now, working on String Theory has advanced math by quite a bit. A lot of great discoveries have come from it, but none are related to the theory itself. It's certainly good people are working on it, but it's likely a dead end as fart as physics goes. As all things, this could change, but that's the state as I am aware of it currently.


*

Online Tumeni

  • *
  • Posts: 3080
    • View Profile
Re: Found a fully working flat earth model?
« Reply #204 on: February 17, 2022, 02:56:06 PM »
Being "inside" a coordinate system is a meaningless concept.
Why are coordinate systems used for travel?

Most county road numbering systems in the US specifically utilize plane geometry, commencing from county center and measured outward in .25 mile increments. North/South roads are oriented East/West and vice - versa.

Are you suggesting that this is a co-ordinate system?
=============================
Not Flat. Happy to prove this, if you ask me.
=============================

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

Nearly?

*

Offline WTF_Seriously

  • *
  • Posts: 1189
  • Nobody Important
    • View Profile
Re: Found a fully working flat earth model?
« Reply #205 on: February 17, 2022, 03:22:45 PM »
It's not a theory. It's a model. It's quite literally just a coordinate transformation.

What do you think is the point of performing this transformation?  I've asked the OP, so what do you think is being achieved here?
I don't think that particularly matters, frankly. Aside from the entertainment value that Pete already mentioned, it may very well not have many practical applications (hence, the phrase "logically sound but practically useless" or however it was phrased earlier). But lots of things start out that way, especially in mathematics. String theory was considered useless to the physics world shortly after its creation, and it was only after many years that people started revisiting it and finding it might possibly have some practical applications.

The truly interesting thing isn't the model OP is sharing. It's the visceral response from what I guess I can only describe as RE ideologues that's really interesting. Some people are so caught up in fighting against a thing, that they can't even realize when they start unknowingly arguing against their own position.

If I had to guess, OP is studying some things like mapping and/or coordinate transformations presently and found an interesting way to take a deeper dive into the topic than just reading stale textbook material. And that's a perfectly valid reason to want to work through a problem like this as well.

The occasional thread like this is one of the things worth sticking around this site for.  When one comes around that is a little educational it's quite refreshing.
Flat-Earthers seem to have a very low standard of evidence for what they want to believe but an impossibly high standard of evidence for what they don’t want to believe.

Lee McIntyre, Boston University

Offline Action80

  • *
  • Posts: 2006
    • View Profile
Re: Found a fully working flat earth model?
« Reply #206 on: February 17, 2022, 05:33:40 PM »
Being "inside" a coordinate system is a meaningless concept.
Why are coordinate systems used for travel?

Most county road numbering systems in the US specifically utilize plane geometry, commencing from county center and measured outward in .25 mile increments. North/South roads are oriented East/West and vice - versa.

Are you suggesting that this is a co-ordinate system?
Of course it is a coordinate system.
It's so hard to have faith in humanity when they do shit like this.

"I hate the police so I'm gonna burn a Walgreen's!"

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #207 on: February 17, 2022, 09:24:20 PM »
Why are you doing this? You said you've run it past graduate physicists or similar, and they pronounced it "practically useless", so what's your purpose?
i believe it's very relevant for the flat-earth "debate".
I find that people are unreasonably attached to shape.
It basically makes the entire debate as moot as polar vs cartesian coordinates.
The simple truth is you can't know the shape of the earth. One person can claim it's flat with bendy light, the next may claim it's a globe.
It's just a different view. It's like 2 colorblind people both insisting their color is the right one....

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #208 on: February 17, 2022, 09:32:59 PM »
The map is exactly as accurate as it's the same physics.
Take a logarithmic X and Y axis and draw a circle around (1000,1000) with diameter 1000. I will not longer look like circle.
Are logarithmic scales therefore "wrong" and need to be "reexamined"?

How do you measure a distance on a globe scale-model?
- You take ruler and measure the distance. (Incidentally note rulers don't fit to globes very well. Ideally you'd need a curvy ruler)
- You calculate the distance using math (multiplying by the scale factor)

You measure distances on the flat-earth representation of physics exactly the same way. (Do note that distances are not scale and rotation invariant so you might find this more akin to finding distances on a logarithmic scale)
- you find start and end point
- you calculate the distance using math. (eg haversine)
Do note, in practice i very seldomly see people calculating distances using globes. Most people just use coordinates and maths.

Okay, so what does your map report as the distances from New York to San Francisco then San Francisco to Brisbane then Brisbane to Perth?
My model is in celestial coordinates (latitude, longitude, distance from the center of the earth)
And as all cities have a fixed distance of 6000km we can simply use the haversine formula.
So to know the distance between cities:
- First, convert all cities to (latitude, longitude).
- Then use haversine to find the distances.
It's exactly the same formula as on a globe, so the result will also be identical.
(lat, lon) are invariant to AE transformation and so the formula is also invariant.

Or in general distance_in_celestial(p1, p2) = distance_in_cartesian(celest_to_cart(p1), celest_to_cart(p2))
So by definition, it will always produce the same distance result as on a globe.
It's just a different representation. Same physics...

*

Offline WTF_Seriously

  • *
  • Posts: 1189
  • Nobody Important
    • View Profile
Re: Found a fully working flat earth model?
« Reply #209 on: February 17, 2022, 09:40:32 PM »
And as all cities have a fixed distance of 6000km we can simply use the haversine formula.

This wouldn't be the case on a flat disc, would it?
Flat-Earthers seem to have a very low standard of evidence for what they want to believe but an impossibly high standard of evidence for what they don’t want to believe.

Lee McIntyre, Boston University

Offline Rog

  • *
  • Posts: 69
    • View Profile
Re: Found a fully working flat earth model?
« Reply #210 on: February 17, 2022, 09:42:41 PM »
Quote
If I understand the model, and Troolon, correct me if I'm wrong, I think it's pretty simple. It's a great circle, identical in both the RE and Troolon models. Same route, same distance. Just depicted, portrayed, visually different. Just like how the route would show as curved on a Mercator projection though it appears "straight" on a globe.

A Mercator projection isn’t making the claim that the earth is actually flat. It shows a curved route because that represents the shortest distance on a globe.

If the earth were actually flat the shortest distance between the two points would be a straight line and different route. A route from NY to Moscow, passing the tip of Greenland and coming back down through northern Europe isn’t the shortest distance on a flat earth because it isn’t a straight line.



If there were no distortions in Trooloon’s model, if all the distances and angles were same, the shortest distance would be the same route because all the land masses would be in the same relative position and have the same relative distance between them.
« Last Edit: February 17, 2022, 09:45:09 PM by Rog »

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #211 on: February 17, 2022, 09:44:06 PM »
I suspect there's a pesky singularity if you go below the surface far enough, though ;)
I haven't encountered a pasky singularity yet: no divisions by zero.
But you're right there's some unintuitive behavior at the center of the earth, but nothing worse than expressing cartesian (0,0) in polar coordinates :)

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #212 on: February 17, 2022, 10:16:56 PM »
Not treating it as any coordinates.  Treating it as simply distance between two points.  Distance can be measured and distances must equate if the two models are indeed similar.

In the RE model we can call distance 'Rounds'  In yours we can call them 'Troolons'.  Regardless the coordinate system, the distance from A to B is X and the distance from A to C is Y. The value of Rounds and Troolons need not equate but the distance in both models must still be the same.  In addition, there must be a conversion factor that is constant between the two models for converting one distance to another.
- a straight ruler as you know it, is only translation-, rotation-, reflection-invariant in an orthonormal base. It also only scales linearly as long as you stay in a orthonormal base.
- If you really want to measure distances in the flat-world representation you should use rather special curvy rulers and be aware that they're not translation or rotation invariant.
  However if i were to take by bendy flat-earth ruler to your globe all distances would also be totally "broken".
- there are multiple possible distance metrics. You already mentioned two: spherical distance and euclidean distance, but there are infinitely many more distance metrics and most of them are not measurable with a ruler.

I will agree that rulers behaving well is a very handy property of orthonormal bases.
But what you're really doing here is picking one very specific shaped ruler, and one very specific base that accidentally work nice together, and therefore declare this the only valid way to measure. Or actually worse, you take your "straight" ruler to my non-orthonormal base and just declare it broken.

Quote
Looking at your map, if Sydney to LAX is 1 troolon, it appears that Sydney to Santiago is on the order of 1.5 troolons.  For the two models to be identical it should be .94 troolons.  Admittedly, I may be misunderstanding how things are measured in your system but your flat disc map seems pretty easy to interpret with respect to relative distances.
My coordinates are in (lat, long). So my distances are calculated with haversine, and will thus give the same result as on your globe.
You're taking a straight ruler, to a non-orthonormal base and then applying the linear scaling property of straight-rulers in orthonormal bases.
Math says no :)

Or if you like a less mathy analogy: You can measure bar-charts with a ruler, linearly scale it and find the value it represents.
You can't do this with a pie-chart. That doesn't mean pie-charts are wrong or broken.
They're just a different way to view the same data.

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #213 on: February 17, 2022, 10:42:20 PM »
If it is impossible for us to 'step outside' to see the true shape of things then there is no way of knowing if your theory is true or not.
If something is impossible to observe then there is no way to prove it exists one way or another.  If you can't disprove a theory, it's not a theory.
Very correct.
The only problem is that everything you've posited above is also true for orthonormal bases and globe earth.
So globe shape is also an unfalsifiable theory.
One minor detail i would like to clarify is that it's impossible to see a difference between the flat and globe representation of physics.
We can of course make measurements in reality and check that they align with what physics predicts in both models.
It's just the shape we can't differentiate.
Orthornormal bases have some nice properties, and we're so very used to them that we sometimes forget it's just a convention.

Quote
We can only know what we can see and measure. In my experience, my measurements and observations show the Earth to be a globe. If it's another shape in a hypothetical greater universe that I can't see or touch or examine in any way, then that's the realm of religion, belief and faith.
I've just created a second model, that also explains all your experiences, measurements and observations equally well on a universe with a flat earth.
It's the same physics, just a different shape. Like bar charts and pie charts.... Math can't see shape, only numbers and relations between them.
So globe shape is indeed in the realm of speculation.
That's indeed the point i was making.
Do note it also works the other way around. If someone ever makes a flat-earth model, i'll use the same trick to turn it into a globe.

*

Offline JSS

  • *
  • Posts: 1618
  • Math is math!
    • View Profile
Re: Found a fully working flat earth model?
« Reply #214 on: February 17, 2022, 10:57:02 PM »
If it is impossible for us to 'step outside' to see the true shape of things then there is no way of knowing if your theory is true or not.
If something is impossible to observe then there is no way to prove it exists one way or another.  If you can't disprove a theory, it's not a theory.
Very correct.
The only problem is that everything you've posited above is also true for orthonormal bases and globe earth.
So globe shape is also an unfalsifiable theory.
One minor detail i would like to clarify is that it's impossible to see a difference between the flat and globe representation of physics.
We can of course make measurements in reality and check that they align with what physics predicts in both models.
It's just the shape we can't differentiate.
Orthornormal bases have some nice properties, and we're so very used to them that we sometimes forget it's just a convention.

The difference is we can see the Earth. We can measure it. We can run experiments on and around it.

It's not the same as some otherworldly force or the universe being a simulation. Those are untestable and unverifiable.

Globe Earth can be disproved.  Finding a dome.  Locating the edge.  Drilling through to the underside.  Seeing it from high enough to see the entire earth as a disk. Plotting the locations of cities and finding them to align on a flat plane but not a sphere.  Exploring past the ice wall to find an infinite plane. 

Quote
We can only know what we can see and measure. In my experience, my measurements and observations show the Earth to be a globe. If it's another shape in a hypothetical greater universe that I can't see or touch or examine in any way, then that's the realm of religion, belief and faith.
I've just created a second model, that also explains all your experiences, measurements and observations equally well on a universe with a flat earth.
It's the same physics, just a different shape. Like bar charts and pie charts.... Math can't see shape, only numbers and relations between them.
So globe shape is indeed in the realm of speculation.
That's indeed the point i was making.
Do note it also works the other way around. If someone ever makes a flat-earth model, i'll use the same trick to turn it into a globe.

You can't simply apply a mathematical transformation to an object and claim that the object is now transformed.  I could write a formula to transform the Earth into a cube shape and render it, but it doesn't make it real.  I can multiply my height by a number and say I'm 100 feet tall and say that is my true height and shape, but that doesn't change my actual height.

I can say latitude ' = latitude x 2 and now have an Earth twice the size.  But it doesn't mean anything, even if I render a stretched Earth, flat or round.

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #215 on: February 18, 2022, 12:11:21 AM »
If I had to guess, OP is studying some things like mapping and/or coordinate transformations presently and found an interesting way to take a deeper dive into the topic than just reading stale textbook material. And that's a perfectly valid reason to want to work through a problem like this as well.
Actually, 4 weeks ago i was puzzled that there had been so little improvement in the flat earth model over the years that i decided to investigate myself. At this point my take was that flat earth is obviously wrong, but there must be something better than the fully illuminated flat disc that's shown in all the media.
So I started very easy by adding disappearing keels to the flat map and discovered the flat earth horizon formula. When you then realize you can't see below the horizon in reality and that the horizon on a flat map is a cone you can't see outside of, then all of optics suddenly starts to work (ie day/night/seasons/....)
But i started running into mathematical difficulties modeling this and so I stumbled upon the coord transform learning that flat-earth and globe-earth physics are really the same thing.

I then went through a brief period of: "have i finally gone totally bonkers, or just the regular nutty with a hint of inspiration"
So that's when I had my results double-checked and entered a bit of a moral dilemma whether to make this public or not.
In the end i chose to because:
- someone else is bound to find it (bendy light theory came sooo close)
- the science community might be briefly surprised, but science adapts and so i doubt it would do any long-term damage
- I find the insults and dogma on both sides of the debate very irritating. I'm hoping this will tone both sides down and hopefully bring them together.

I don't get any enjoyment of riling people up, quite the contrary.
The biggest sentiment i get from this is frustration with myself at not being able to communicate this more clearly.

Though i must say i'm extremely thankful everyone has remained polite and some of the discussions have been very interesting.
And if some bozo came to me claiming we couldn't ever know the shape of the planet, i would be equally incredulous and eager to prove him wrong :)
So thank you everyone for the exchanges so far.

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #216 on: February 18, 2022, 10:42:54 PM »
Quote
Straight is only defined the way you think it is in an orthonormal basis.
Straight is the shortest distance between two points, regardless of the coordinate system.  Straight is a curve on a sphere.  What would be the shortest distance between NY and Moscow on your model?
[/quote]
After coord transform we're working in (lat,long) based coordinates. So haversine gives the shortest distance for 2 points on earth.

Quote
Not all transformations are isometric, which just means a transformation that it doesn’t distort angles or distances.
Here is the mathematical proof. (starts around slide 20) I can’t help you much working through it, but I suggest that unless you can contradict the math, you really don’t have any basis to say that your model doesn’t have any distortions.
The problem in the mathematical proof is on slide 5:
     We imagine trying to draw a map of a region D ⊂ S2. The map is a smooth function  F : D → E 2
     to the Euclidean plane that preserves as much information about the geometry of D as is possible.
I am not working in a euclidean plane. After coords transform my basis is not orthonormal.

I've told you this before: you're drawing an orthonormal basis next to mine and then start measuring things according to your base. That's now how coord transforms, maths or physics work.

Also i found this definition for isometry: https://en.wikipedia.org/wiki/Isometry.
It's basically any transformation that preserves distances. And my transform preserves distances. It is isometric.
What you're doing is taking a coord transformed base, putting an orthonormal basis next to it, and then start measuring. That's not the definition of isometric. You have to use the distance metric of your coord transfromed metric space.

Quote
Since the curvature of the sphere does not vanish, it CANNOT BE LOCALLY ISOMETRICALLY MAPPED TO THE EUCLIDEAN PLANE
except my "disc" is not an euclidean plane. It does not have an orthnormal base.
- You've had me calculate curvature of my space using gauss and it came out indistinguishable from a sphere.
- You then had me do parallel transports, which were indistinguishable from a sphere.
- I've given you the calculation for riemann tensor and ricci scalar, and it was comparable with a sphere.
You never found fault in any of the math, and yet you now call it a euclidean plane which violates all of the above.

But here's a very simple visual example: Take a compass and draw a circle anywhere on the AE map.
Circles represent all points at the same distance from the center. Does this circle represent all points at equal distance?
Obviously not, you said yourself there's distortion.
So my "disc" is demonstrably not an euclidean geometry.

Quote
At what point have i tried to "flatten" the sphere?
-> By trying to project a sphere onto a flat surface without any distortion.
Not treating it as a euclidean geometry.

Quote
If it has intrinsic curvature, it doesn't matter how you view it.  It will always have intrinsic curvature and you can always detect it with some simple tests.
Agreed. Mathematically it behaves like a euclidean globe. But i can represent it whatever way i like.
I can represent it in an orthonormal base and say i live on a sphere with straight light.
Or i can represent it in my way, and look at reality as a flat earth with curvy light.
I'm able to construct a consistent world view either way and i'm able to make the same predictions of reality in either representation.
However when i switch to FE representation, i should be aware not to use any properties exclusive to euclidean geometries.



Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #217 on: February 18, 2022, 10:45:01 PM »
Quote
Straight is the shortest distance between two points, regardless of the coordinate system.  Straight is a curve on a sphere.  What would be the shortest distance between NY and Moscow on your model?

If I understand the model, and Troolon, correct me if I'm wrong, I think it's pretty simple. It's a great circle, identical in both the RE and Troolon models. Same route, same distance. Just depicted, portrayed, visually different. Just like how the route would show as curved on a Mercator projection though it appears "straight" on a globe.
Thumbs up, you've got it :)

Offline Rog

  • *
  • Posts: 69
    • View Profile
Re: Found a fully working flat earth model?
« Reply #218 on: February 19, 2022, 09:49:32 AM »
Quote
except my "disc" is not an euclidean plane. It does not have an orthnormal base.
- You've had me calculate curvature of my space using gauss and it came out indistinguishable from a sphere.
- You then had me do parallel transports, which were indistinguishable from a sphere.
- I've given you the calculation for riemann tensor and ricci scalar, and it was comparable with a sphere.
You never found fault in any of the math, and yet you now call it a euclidean plane which violates all of the above 

It doesn’t matter if your model has an orthonormal basis or not. That isn’t what defines euclidean space.   Where have you posted a graphic of a parallel transport of your model?  How would it differ from the example I posted?  Nor have I seen any calculations of what the Gaussian curvature of your model is or the calculations of the Riemann tensor.  A few days ago you acknowledged that much of the math is over your head.  Now you’re an expert in differential geometry and tensor calculus?

The Riemann tensor is invariant.  If it exists in one coordinate system, it exists in every coordinate system. If it is zero in one coordinate, it is zero in all of them.

Quote
This tensor DOES NOT CARE about coordinate systems. Flat space has all components of Ra bcd = 0 irrespective of whether you are working in spherical polar coordinates or cartesian coordinates

http://astro.dur.ac.uk/~done/gr/l9.pdf

Quote
In other words, the independent components of the Riemann tensor can be thought of as the n 2 (n 2 − 1)/12 (linear combinations) of second derivatives of the metric tensor that cannot be set to zero by coordinate transformations.

https://cosmo.nyu.edu/yacine/teaching/GR_2019/lectures/lecture11.pdf

Quote
But because the Riemann tensor is a genuine tensor, if it vanishes in one coordinate system then it must vanishes in all of them.

http://www.damtp.cam.ac.uk/user/tong/gr/three.pdf

Quote
Is it possible to find an inertial coordinate system such that also the second derivatives of the metric tensor vanish at the coordinates origin? The answer is negative: the reason being that the Riemann curvature tensor (see Chapter 5) cannot be made to vanish by use of coordinate transformations.

https://arxiv.org/pdf/1509.01243.pdf

Quote
Proof: if a tensor is zero in any frame, then it is zero in all frames, as a trivial consequence of the transformation law 

https://www2.physics.ox.ac.uk/sites/default/files/2012-09-20/gr_b5_pdf_20413.pdf

Do I really need to keep going?


Quote
 Mathematically it behaves like a euclidean globe.


There’s no such thing as a euclidean globe. The definition of a globe is that it doesn’t “behave”, or have the all of the same properties of a euclidean space

Quote
We imagine trying to draw a map of a region D ⊂ S2. The map is a smooth function  F : D → E 2
     to the Euclidean plane that preserves as much information about the geometry of D as is possible.
I am not working in a euclidean plane. After coords transform my basis is not orthonormal.

One more time...it doesn’t matter if the basis isn’t orthonormal.  If it doesn’t have curvature, it is a euclidean plane.  If it has curvature it is a sphere.  Coordinate transformations don’t make any difference.

Quote
The metric is flat if the Riemann curvature tensor is zero. That's true regardless of what coordinates you use. Spherical coordinates can be used in a flat space, just as polar coordinates can be used on a flat plane.

https://www.physicsforums.com/threads/spherical-coordinates-metric.761459/


Quote
Also i found this definition for isometry: https://en.wikipedia.org/wiki/Isometry.
It's basically any transformation that preserves distance

It doesn’t just preserve distance.  It preserves angles.  You can’t have an isometric transformation between a plane and a sphere because the angles of a plane and a sphere are necessarily different. See below…

Quote
My coordinates are in (lat, long). So my distances are calculated with haversine, and will thus give the same result as on your globe.

No they won’t.  Haversine calculates great circle distances.  If you connect three great circles to make a triangle on a sphere, it will have more than 180 degrees.  You can’t have a triangle of more than 180 degrees on a flat surface.  So the area represented by connecting three “great circles” on a flat disc will be different than the area represented by connecting the same three “great circles” on a sphere.

The supposedly same three points won’t meet in the same place on a plane and sphere and the angles connecting the supposedly same three points are different.

Offline troolon

  • *
  • Posts: 101
    • View Profile
Re: Found a fully working flat earth model?
« Reply #219 on: February 21, 2022, 10:37:44 PM »
Also, this is wrong. As it has been mentioned, curvature is invariant over a change of coordinates. If it's non-zero, there is no possible change of coordinates that can turn it to zero globally.
Doesn't matter the axis or coordinate basis. If your model has non-zero intrinsic curvature, it isn't flat
You're quite right the intrinsic curvature is indeed curved. Mathematically it's a curved space.
But i believe extrinsic curvature to be zero (i'm sure Rog will have a 70 page proof or something about this :)

When you look at a model of the world, you look at the model from the outside. So i believe it's correct to say it's a flat representation of physics.

For me personally i can even look at the world and experience it as the flat representation.
For example when i see a picture of a ship with a missing keel, i can very often imagine it's the light that's curving.
Or when i see a globe from space, i can superimpose the model and tell myself: ah yes, that's a flat disc distorted by curvy light.

Some people seem to feel more comfortable looking at the world as flat, and now science can explain the world in their view if they want to...