Re: Found a fully working flat earth model?
« Reply #220 on: February 22, 2022, 07:28:16 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...

The extrinsic curvature is only defined when you embed a submanifold in a higher dimensional one. So for us it does not matter, since we can only experience 3 space dimensions; if the earth has zero extrinsic curvature for some beings that can experience 4D space, whatever. But even they will agree that for us mere 3D beings, earth is curved. So your model is curved (where it matters), not flat.

To the second part: of course you can assume earth is any shape you like and imagine light does whatever to conform to that. But then you're taking the shape as a premise, not conclusion. And "science" cannot explain a flat earth, since by definition it has to invoke yet unexplained physics, such as the behaviour of light. Which has never been seen to do what you want it for earth to be flat, and furthermore has no explanation at all beyond "light curves in the exact way for earth to be seen as a globe from space, for us to see sunsets and sunrises, for us to see objects with their bottom part obscured...". If you want to call this a "model" sure, but I think most of us expect a little more. And this is certainly not science

Offline Gonzo

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Re: Found a fully working flat earth model?
« Reply #221 on: February 22, 2022, 11:38:36 PM »
It’s certainly an interesting thought experiment, but agreed, it’s not a model, let alone a ‘fully working’ model.

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Offline Clyde Frog

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Re: Found a fully working flat earth model?
« Reply #222 on: February 23, 2022, 01:28:25 AM »
https://www.education.vic.gov.au/school/teachers/teachingresources/discipline/science/continuum/Pages/scimodels.aspx
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A scientific model is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted. The models scientists create need to be consistent with our observations, inferences and current explanations. However, scientific models are not created to be factual statements about the world.

It's... not a model?

Offline Rog

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Re: Found a fully working flat earth model?
« Reply #223 on: February 23, 2022, 02:13:11 AM »
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It’s certainly an interesting thought experiment, but agreed, it’s not a model, let alone a ‘fully working’ model.

OP seems to suffer from the same misconception that many have on this site.  That there is something magical about changing coordinates or frames of reference and that it actually effects or changes physical reality. It doesn’t.

The idea that we can never know the true shape of the earth, or that it could be anything, starts with the assumption that there aren’t physical forces at work that determine the shape.  Everything is perception and one perception is as valid as another.  That is such bunk. Perception is not always reality and there are ways to test perception against reality

Re: Found a fully working flat earth model?
« Reply #224 on: February 23, 2022, 01:18:36 PM »
https://www.education.vic.gov.au/school/teachers/teachingresources/discipline/science/continuum/Pages/scimodels.aspx
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A scientific model is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted. The models scientists create need to be consistent with our observations, inferences and current explanations. However, scientific models are not created to be factual statements about the world.

It's... not a model?

OP's model is a model of the round earth, agreed. But it's not a model of a flat earth according to the quote, since there's no current explanation for bendy light (or light behaviour in general).

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Offline Clyde Frog

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Re: Found a fully working flat earth model?
« Reply #225 on: February 23, 2022, 01:51:54 PM »
https://www.education.vic.gov.au/school/teachers/teachingresources/discipline/science/continuum/Pages/scimodels.aspx
Quote
A scientific model is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted. The models scientists create need to be consistent with our observations, inferences and current explanations. However, scientific models are not created to be factual statements about the world.

It's... not a model?

OP's model is a model of the round earth, agreed. But it's not a model of a flat earth according to the quote, since there's no current explanation for bendy light (or light behaviour in general).
It doesn't need to explain that though. It just needs to ensure light's behavior is effectively modeled. The clue is in the name.

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #226 on: February 23, 2022, 10:06:33 PM »
UCLA Math/Computer Science 1975. Never used the math because I became a software engineer. I can barely remember the math, but I learned stuff like what a coordinate system is and can refresh my math memory with internet. I remember enough to see the wrong math you do when make it complicated in order to hide the wrong part. Arguing details is pointless, you are not here to learn, you are here to justify FE.

I talked about FE with a programmer friend. He said that he had written the nav software for the Canadian Air Traffic Control system and used 3d geometry, haversines, straight out of a math textbook. Worked perfectly, airplanes in Canada today arrive exactly where they are supposed to, FE math would be different and not work. He has a degree in math.

I would be astounded if there was a qualified mathematician or physicist that agreed with any of Troolon's math. I am not surprised that he cites secret experts. Confirmation without possibility of falsification.

I remember when my linear algebra prof started the lecture with "Today we are going to talk about how we know the earth is not flat, Gauss' Remarkable Theorem." The normal vectors on the surface of a flat disk are parallel, so curvature is zero. The normal vectors on the surface of of a sphere are not parallel, so the curvature is not zero. Find me a math prof who will disagree with this. There aren't any. There is a reason why you are sayoing this on FE site, say it on a math web site, or astronomy, or astrophysics.

What you did is simple, you peeled the surface of the sphere off from the south pole and stretched it out into a disk, like popping a balloon and stretching the balloon out into a disk. All the rest is just blather. Doing this stretches out the size of everything in the southern hemisphere. In your graphics, Australia is clearly bigger than North America. Measurement is measurement, and there are many ways to know the true sizes.

But distance is not your only problem. When you stretch it, the south pole becomes the circle around the edge. If you incorporate Sigma Octantus, as directly above south pole, it becomes a circle rather than a particular point. Even if this made sense, you have to explain why observers in the southern hemisphere see it as a small dot on the part of the circle and do not see the rest of the circle. A difficult to explain combination of bent light and directional light, much like the spotlight sun problem, but worse.

Except for one thing, Sigma Octantus is not directly above the south pole, it is a little over a degree off axis. So consider the Southern Cross. It is enough south that it is seen from everywhere in the southern hemisphere as being due south. It is much like the big and little dippers in northern hemisphere. Where is the Southern Cross? It appears everywhere as outward from the disk. Where is it really? That question has no sensible answer on FE.



According to your theory, since everything is equivalent, seems like I should be able to do the same math conversion using the south pole. This gives Sigma Octantus pretty close to observed re azimuth. still requires bent light ("unknown forces with unknown equations", per the FAQ) for altitude. But now Polaris, the north star, is everywhere around the edge. Also seems pretty arbitrary to start with the poles, how about your house in the center? Start with your house and peel the surface of the sphere starting with the point directly opposite on the spherical globe. The same transform can be done choosing any point at the center. If you choose a pole, one of the pole stars makes at least some sense. Start at the equator and neither makes any sense at all.

You made graphics to show a transform to several different shapes. Include the pole stars in these transforms and see how much sense it makes. Let's see graphics that include the Southern Cross and the Little Dipper, visible from the places and in the direction that matches observations. You can't do it. Distances, direction, and observed location of astronomical objects are all correct in RE graphics. They are not correct in any other shape.

"Electromagnetic Acceleration" sounds so much more sciency than "bendy light".

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #227 on: February 25, 2022, 08:25:36 PM »
... we can simply use the haversine formula.

This wouldn't be the case on a flat disc, would it?

You can have infinitely many distance metrics. You have to choose one that make sense and haversine seems to match reality.
Intrinsically my model behaves like a globe. Extrinsically it's flat. When we look at a model, we look at it from the outside so extrinsically.
So i would argue it's possible to create a flat earth representation, that mathematically behaves just like a globe would.

I think the entire dialog can be split into 3 questions:
- What does the earth look like in an orthonormal reference frame/euclidian geometry
    -> a sphere / obloid spheroid.
- Can we model physics on a flat earth?
    -> yes, my model does that.
- What is the real shape of the planet / what is the correct coordinate system?
    -> coordinate system is a choice not a given especially for the universe that we can't observe externally.

I believe when flatties and globies are debating shape, they're just discussing coordinate systems. (unless they make additional claims)

Re: Found a fully working flat earth model?
« Reply #228 on: February 25, 2022, 09:16:10 PM »
Quote from: troolon
You can have infinitely many distance metrics. You have to choose one that make sense and haversine seems to match reality.
Yes, but a flat surface, by definition, has an euclidian metric. That is, the euclidian metric gives the shortest distance between 2 points, not haversine.

Quote from: troolon
Intrinsically my model behaves like a globe. Extrinsically it's flat. When we look at a model, we look at it from the outside so extrinsically.
So i would argue it's possible to create a flat earth representation, that mathematically behaves just like a globe would.

There seems to be a bit of confusion here. Extrinsic curvature has a very precise definition: it's the curvature seen by embedding the surface in a higher dimensional manifold. In order to see the extrinsic curvature of a 3-sphere, you'd need to be in 4 spatial dimensions. We are not, therefore we cannot see the extrinsic curvature in reality. You could, if you embed your earth in R4, but that's a purely mathematical tool that has no correspondence in reality since we live in 3 spatial dimensions. So in this sense you could "look at the model from outside", with the caveat that this cannot be done in the real world. So if you claim that earth is flat in 4D space, whatever. It's untestable and unfalsifiable, specially because you said if behaves exactly like a globe in 3D space. Furthermore, for the inhabitants of this world, it would appear curved since the intrinsic curvature is not zero.

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- What does the earth look like in an orthonormal reference frame/euclidian geometry
This question is a bit strange. We have mentioned this before, but curvature does not care about coordinates. I don't understand how you can look at a surface in "euclidian geometry". Its properties are not dependent on the coordinate system you use. And by the way, a (lon,lat) system of coordinates on a sphere is also orthonormal, as are polar coordinates in a plane, cylindrical coordinates, ...

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- Can we model physics on a flat earth?
You have yet to show that since your model is not flat, unless in the sense described above.

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- What is the real shape of the planet / what is the correct coordinate system?
These questions are completely unrelated, and there's no correct coordinate system. You can choose whatever you want, but as we said before, the properties of the surface such as curvature are not coordinate dependent.

Offline Rog

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Re: Found a fully working flat earth model?
« Reply #229 on: February 25, 2022, 11:51:46 PM »
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You can have infinitely many distance metrics. You have to choose one that make sense and haversine seems to match reality

The reason Haversine makes sense and matches reality is because Haversine formula gives you the shortest distance between two points on a sphere. It is specifically designed to measure the distance on a sphere. If you want to find the shortest distance between two points on a plane, you have to use the PT.  That is the formula that will give you the shortest distance between points on a plane.  If you use the same two points in the two different formulas, you will get different distances, because distances on a sphere are measured differently than on a plane.

You seem to lack the fundamental understanding that Euclidean and non-Euclidean geometry are different.  Different rules, different definitions, different ways of measuring. 

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #230 on: February 25, 2022, 11:54:27 PM »
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?
1. Is it possible to make physics work on a flat model? ie can the flatties ever find a working flat earth model?
  -> yes. Of course as both globe and flat representation represent the same reality, it would be very surprising if there were no correlation between the two models.
2. What's the outside shape of the universe?
  -> We can't know



Offline troolon

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Re: Found a fully working flat earth model?
« Reply #231 on: February 26, 2022, 12:17:37 AM »
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.
There are loads of different distances. Straight through the earth, angular distance, taxicab. hamming distance if you like a weird one.
It's your responsibility to take one that makes sense: that matches observations.
Personally i find it remarkable that physics uses both spherical and euclidean distances within the same model. This alone clearly shows there's no one correct distance per model.

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Offline WTF_Seriously

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Re: Found a fully working flat earth model?
« Reply #232 on: February 26, 2022, 04:11:40 PM »
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?


This is not my quote.  If you're going to reference quotes make sure they are attributed to the correct source.
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

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Re: Found a fully working flat earth model?
« Reply #233 on: February 26, 2022, 05:09:00 PM »
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There are loads of different distances. Straight through the earth, angular distance, taxicab. hamming distance if you like a weird one.
It's your responsibility to take one that makes sense: that matches observations. Personally i find it remarkable that physics uses both spherical and euclidean distances within the same model. This alone clearly shows there's no one correct distance per model

There is only one correct shortest distance between two points per model because you have to use different formulas for each.  Haversine will give you the shortest distance between two points on sphere and the PT will give you the shortest distance between points on a plane. 

The taxicab metric assumes right angles are necessary to travel from point A to point B.  So it doesn’t make sense to use it when that’s not necessary. Haversine assumes an angle between the two points created by the curvature of the surface of a sphere. On a plane, so such angle is created because there is no curvature, so it doesn’t make sense to use Haversine on a plane.  But still you insist on doing so.



Those are the distances measured  between Perth and Johannesburg using the correct formulas for each model.

http://walter.bislins.ch/bloge/index.asp?page=Distances+on+Globe+and+Flat+Earth

When you use the correct formula for the correct model, you get two different distances, and it’s easy enough to determine which one reflects reality.


Re: Found a fully working flat earth model?
« Reply #234 on: February 27, 2022, 08:24:47 PM »
I believe to have found a fully working flat earth model. Anything that can be proven by physics can also be proven in it.
It's very similar to the bendy light/electromagnetic acceleration theory.
All details are on my website including animations of day/night/seasons: https://troolon.com.
But yes, i believe a working flat earth model has finally been developed.

Feel free to have a look.
Troolon
The term "flat earth" as used in your posts is not the same as the that same term as used by FE claimants on this site (i.e. a observable flat earth in the environment in which the earth and we exist).  Thus you do NOT have a working flat earth model.

Physics is not math.  The relationships between dimension, space, mass, energy, etc that are found in physics can (so far) be represented in math.  That in no way implies that any math (especially just for the surface of objects) represents a potential physical reality.
« Last Edit: February 28, 2022, 01:10:24 AM by ichoosereality »
If "bendy light" were real the spot shape and power output of large solid-state lasers would vary depending on their orientation relative to the surface of the earth, but this is not observed thus bendy light is not real.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #235 on: March 02, 2022, 09:54:49 PM »
you are here to justify FE.
I'm here to show
- It is possible to do physics with a flat representation of earth
- We can't know the true shape of the planet as observed by an external observer
- the RE/FE discussion is just as pointless as polar vs cartesian coords.

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I talked about FE with a programmer friend. He said that he had written the nav software for the Canadian Air Traffic Control system and used 3d geometry, haversines, straight out of a math textbook. Worked perfectly, airplanes in Canada today arrive exactly where they are supposed to, FE math would be different and not work. He has a degree in math.
Totally agree. I never claimed RE is wrong. FE and RE are just two representations of reality. You can write software using either one (or even switch back and forth if you iike)

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I remember when my linear algebra prof started the lecture with "Today we are going to talk about how we know the earth is not flat, Gauss' Remarkable Theorem." The normal vectors on the surface of a flat disk are parallel, so curvature is zero. The normal vectors on the surface of of a sphere are not parallel, so the curvature is not zero. Find me a math prof who will disagree with this. There aren't any. There is a reason why you are sayoing this on FE site, say it on a math web site, or astronomy, or astrophysics.
and i agree with your prof, yet i can represent all of physics on a flat plane and have it working. It's intrinsic versus extrinsic curvature.

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What you did is simple, you peeled the surface of the sphere off from the south pole and stretched it out into a disk, like popping a balloon and stretching the balloon out into a disk. All the rest is just blather. Doing this stretches out the size of everything in the southern hemisphere. In your graphics, Australia is clearly bigger than North America. Measurement is measurement, and there are many ways to know the true sizes.
And yet Brisbane and every other city has the same latitude/longitude on both the sphere and the disc.
Lat/long coordinates are invariants. Angle (or angular distance/haversine) is another invariant.
Keeping the invariants in mind, it is possible to do math/physics on a flat disc. In fact i've shown it's possible to do calculate everything physics can.

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But distance is not your only problem. When you stretch it, the south pole becomes the circle around the edge. If you incorporate Sigma Octantus, as directly above south pole, it becomes a circle rather than a particular point. Even if this made sense, you have to explain why observers in the southern hemisphere see it as a small dot on the part of the circle and do not see the rest of the circle. A difficult to explain combination of bent light and directional light, much like the spotlight sun problem, but worse.

Except for one thing, Sigma Octantus is not directly above the south pole, it is a little over a degree off axis. So consider the Southern Cross. It is enough south that it is seen from everywhere in the southern hemisphere as being due south. It is much like the big and little dippers in northern hemisphere. Where is the Southern Cross? It appears everywhere as outward from the disk. Where is it really? That question has no sensible answer on FE.
Mathematically it has an answer. It just does not appeal to your intuition (partly because you've been trained in orthogonal coordinate systems since elementary school)

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According to your theory, since everything is equivalent, seems like I should be able to do the same math conversion using the south pole. This gives Sigma Octantus pretty close to observed re azimuth. still requires bent light ("unknown forces with unknown equations", per the FAQ) for altitude. But now Polaris, the north star, is everywhere around the edge. Also seems pretty arbitrary to start with the poles, how about your house in the center? Start with your house and peel the surface of the sphere starting with the point directly opposite on the spherical globe. The same transform can be done choosing any point at the center. If you choose a pole, one of the pole stars makes at least some sense. Start at the equator and neither makes any sense at all.
I totally agree. You can use any point as center and when doing calculations i often translate the disc to where i'm calculating.
It seems you find the southpole makes little intuitive sense (it does work mathematically) so if you plan to do math around the southpole, I would recommend to transform using the southpole as center of the disc. The point you're making now is if globe earth should be rendered with the northpole on top or the some other point. It doesn't matter.

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You made graphics to show a transform to several different shapes. Include the pole stars in these transforms and see how much sense it makes. Let's see graphics that include the Southern Cross and the Little Dipper, visible from the places and in the direction that matches observations. You can't do it. Distances, direction, and observed location of astronomical objects are all correct in RE graphics. They are not correct in any other shape.
If you're a programmer does this help?
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def draw_as_flat(x, y, z):
  lat, long, dist = convert_xyz_to_latlond(x, y, z)
  render_as_disc(lat, long, dist)
This works for every possible point in space. Every point in space can be rendered with the earth being flat.
If you had your friends program, all that needed to be changed was the render function. (or you can go change all formulas to "FE-formulas" if you really want to)
If you want to understand why certain stars are visible, add lines representing rays of light to your globe model, then transform every point using the function above and you'll see the ray becomes curved.

I had this image lying around. Southern cross is in there somewhere, but i think more near the back where it's hard to make out.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #236 on: March 02, 2022, 10:12:57 PM »
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. 
It's just a different representation of physics. As such there's no measurement or observation that will be predicted differently by either model.
- I'm not claiming a dome (in the model with the dome all of space is contained inside the dome, so you're never able to see or reach the dome, only an external observer could tell)
- Mathematically there's no edge in my model. Latitude goes from [0 to π] and is undefined beyond.
- You can't drill through the underside, you'll automatically curve up again :)
- If light curves as in my model, the disc looks like a globe from space (the light curvature was so designed it exactly counteracts the missing curvature)
- Plotting cities will work in lat/long coordinates. (they won't line up visually though, only mathematically)
...

You can't simply apply a mathematical transformation to an object and claim that the object is now transformed.
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.
Still
- We can't know what the earth looks like when observed from outside the universe
- If someone needs a model of earth, there's nothing wrong if she draws a flat disc. All calculations can be done with it.

Personally i think when people are discussing shape of the earth, they're usually talking about the shape of the model.
I think I've always referred to it as FE model of FE representation and I think that term is correct.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #237 on: March 02, 2022, 11:01:25 PM »
Hello Rog,

First of, i don't disagree with any of the math.
I actually think we're pretty close to understanding one another and i think our disagreement might be mostly grammatical in nature.

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?
I've made a serious effort to try learn this topic and I think i have some basic understanding of curvature now. Of course i'm not an expert.
Picture of parallel transport is in Reply #173 https://forum.tfes.org/index.php?topic=19093.msg259064#msg259064 Please enlarge the attached picture, On the small thumbnail it's not very clear i added the extra vectors.
Gauss was calculated by construction
Bee walking a circle was done in my head

I actually think we agree about this model having intrinsic curvature

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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. 
Then i misunderstood why you included the proof. Intrinsically my model is curved.


I've read the rest of your reply, and i think the basis of our disagreement lies in intrinsic versus extrinsic curvature.
I agree that intrinsically a globe has curvature. I will also agree that coordinate transformations don't change intrinsic curvature.
My model has the intrinsic curvature of a globe. (This has been shown at least 3 or 4 different ways)
However what we haven't yet discussed is extrinsic curvature, ie embedding of the surface in a higher dimension.
And in that sense my model is flat.

Models are typically viewed from the outside. We draw them on a piece of paper and look at them as an external observer.
That is why i call this a flat model. I also believe this is how most people view it.
I do understand that intrisically my model is curved and why you wouldn't call it flat.
The correct term is probably an extrinsically flat model of an intrinsically curved surface.
And as we nearly always work on models from the outside i would abbreviate that to a "flat model" but i can see how that would be confusing to someone approaching it mathematically.

Re: Found a fully working flat earth model?
« Reply #238 on: March 03, 2022, 04:48:37 AM »
- Mathematically there's no edge in my model. Latitude goes from [0 to π] and is undefined beyond.
So how are things transported back to 0?  that is physics that does not exist in the globe model so its not just different math.

- If light curves as in my model, the disc looks like a globe from space (the light curvature was so designed it exactly counteracts the missing curvature)
Light does not curve.  If it did, solidstate lasers would not function identically regardless of how they are oriented with respect to teh surface of the earth, but they do.  So light is not bending.  I explained that in this forum but some mod for some unstated reason moved it here https://forum.tfes.org/index.php?topic=19173.0  .

- Plotting cities will work in lat/long coordinates. (they won't line up visually though, only mathematically)
But traveling between them will not be the same, would not take the same time nor use the same fuel nor have the same angular separation when viewed from a 3rd city.  Guess's Theorema Egregium proves you can not do this without distortion as has been pointed out to you many times. 

- We can't know what the earth looks like when observed from outside the universe
There is no "outside" of the universe.  So your statement amounts to "We can't know what the earth looks like when observed from a non existent location".  Even if you consider that to be true its certainly meaningless.   You often conflate the shape of the universe (which might not even have a meaning and certainly not one that I understand) with the shape of the earth, which we know well.   We can measure the shape of the earth and even the shape of space around the earth (Gravity Probe B}, but that is not the shape of the universe.

- If someone needs a model of earth, there's nothing wrong if she draws a flat disc. All calculations can be done with it.
How would the physics of orbital mechanics work with a flat disk?  How does physics in the flat model explain what keeps an orbiting object in that orbit?  That is not just globe earth orbital mechanics with math transformations, that is extremely different physics with all sorts of discontinuities.  Physics is not just about surfaces.
« Last Edit: March 03, 2022, 08:07:12 AM by ichoosereality »
If "bendy light" were real the spot shape and power output of large solid-state lasers would vary depending on their orientation relative to the surface of the earth, but this is not observed thus bendy light is not real.

Offline Rog

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Re: Found a fully working flat earth model?
« Reply #239 on: March 03, 2022, 06:21:18 PM »
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My model has the intrinsic curvature of a globe. (This has been shown at least 3 or 4 different ways)
However what we haven't yet discussed is extrinsic curvature, ie embedding of the surface in a higher dimension.
And in that sense my model is flat.

Gaussian (intrinsic) curvature is quantified as the product of the two principal (extrinsic) curvatures.  If one (as in a cylinder) or both (as in your flat model) extrinsic curvatures are zero, then the intrinsic curvature is zero.

IOW, it's impossible for your model (or anything else) to have intrinsic curvature unless it has extrinsic curvature.