[troolon]

They are only equivalent if the change was isometric.
Flattening a sphere is more than just a coordinate change because it also changes the metric and the Riemann curvature tensor
When you transform a disc to a sphere the metric changes.  The Riemann tensor changes from 0 to non-zero.  Therefore, a disk and a sphere are not mathematically equivalent.

Just for completeness, i've since shown the riemann tensor is non-zero in my space.
So it's probably fair to say the discussion is ongoing.

I'll try to answer your latest post tonight. But i think in summary my reply will be:
A disc in an orthonormal basis, with euclidian distances/angles/vectors, will have a curvature of 0.
A disc in celestial coords, with spherical distances, will have the curvature of a sphere.
I believe you're taking a disc in celestial coords, and expecting it to behave like a disc in an orthonormal basis.
Intrinsic curvature doesn't see the coordinate axis directly, but it does notice distances, angles and tangent planes behaving relative to this axis.

[Dr Van Nostrand]

Distances can be measured.
What you are really saying is that it is impossible for YOU to know the shape of the Earth. I'm guessing you just don't travel a lot.

Problem is my proof is mathematical by nature, so unfortunately  i've proven it's impossible for you to know the shape of the earth too
Physics is a model of reality. It's just a representation. You can have different representations of the same reality.
All i've done is constructed a different representation of physics...
So when you drive your car and measure distances that way, both representations will predict the same answer.
Soo.. now that we have different shaped representations of reality that can't be distinguished by any measurement, how will you tell what shape planet you're truly on?

When a theory in physics doesn't predict the tangible, measurable reality of the world, it's wrong and has to be reexamined. Your map is simply wrong about the relative sizes, shapes and position of the Earth's major continents.


It would be different if your map produced more accurate measurements then a round Earth map but it doesn't. Are you saying it's impossible for us to measure any distances?

[Tumeni]

I'm a little disappointed by this post.

Boo-hoo.

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?

[Pete Svarrior]

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'd speculate that he (and his audience, myself included!) enjoys watching RE'ers expose themselves as completely mathematically illiterate as you lot struggle to process a fairly basic concept.

Does that prove RE false? Of course not. Does it provide a wealth of information on RE'ers' reliability? Hell yeah it does.

[DuncanDoenitz]

I wouldn't call myself illiterate, Pete, but I'm happy to concede that all this coordinate shenanigans, like GR, is well over my head, which is why I tend to stick to concepts I can get my head around, like carpet.  If that gives you any degree of contentment, then, fine.  It doesn't detract from my self esteem.  I have other attributes, which some find endearing, but I'm glad if we've enhanced your life. 

So the OP claims to have produced a working flat earth model; is it any use?  I have a commemorative plate from the wedding of Prince Charles to Lady Diana Spencer.  Does that mean the Prince of Wales is flat?  No, its just a representation of a 3 dimensional entity (or pair of entities, which some might argue are one-dimensional anyway) on a disc. 

And its actually pretty crappy as a plate.  And his shoulders look distorted. 

So, whad'ya think.  Is it a viable model?

[Pete Svarrior]

I have other attributes, which some find endearing, but I'm glad if we've enhanced your life. 

You're right. There's no shame in not being a subject matter expert on every subject. It's the combination of a lack of knowledge combined with extreme confidence that amuses me. And, for what it's worth, I'm not singling you out - it's the general trend over what somehow became a 10-page thread.

So, whad'ya think.  Is it a viable model?

Depends on what you mean by "viable". It's just the RE model, nothing more, nothing less.

[troolon]

When a theory in physics doesn't predict the tangible, measurable reality of the world, it's wrong and has to be reexamined. Your map is simply wrong about the relative sizes, shapes and position of the Earth's major continents.

It would be different if your map produced more accurate measurements then a round Earth map but it doesn't. Are you saying it's impossible for us to measure any distances?

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.

[troolon]

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

My axis are the equator (circular) and the Greenwich greatcircle. What 'straight' path are you taking?
Straight is only defined the way you think it is in an orthonormal basis.
This drawing is in celestial coordinates (lat, lon, distances). Please don't treat it as (x,y,z) cartesian coords because that will indeed break everything.

[Dr Van Nostrand]

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?

[Clyde Frog]

Have you tried reading the very first page? The distances are going to be the same as on a globe. Because beginning at the surface of the Earth and extending outwards, it's the same model. This is a globe still, expressed with a coordinate transformation. It still preserves the same distances and angles we experience every day. Because beginning at the surface of the Earth and extending outwards, it's the same model.

I suspect there's a pesky singularity if you go below the surface far enough, though

[Tumeni]

I very seldomly see people calculating distances using globes. Most people just use coordinates and maths.

I would suggest that the vast majority of physical measurements of our Earth were made before you and I were born (or at least when we were very young).

The generally-accepted first instance of someone in modern times calculating the circumference of Earth was Norwood in the 1600s.

Ordnance Survey started their mapping of the UK in the 1700s, and completed their "retriangulation of Britain" by 1962, when I were just a lad.

If you don't "see" them, perhaps that's because their work was done in the past. 

[troolon]

What is meant is the intrinsic curvature of the space, meaning it is independent of the choice of coordinates. There are clever methods of determining whether and to what extend your space deviates from flat euclidean space, namely Gaussian curvature, and, more importantly, the Riemann tensor.
You are correct that transforming from Cartesian to Geo graphic coordinates doesn’t change the intrinsic curvature, but just drawing latitude on a straight axis doesn’t change it either. 

Would have loved if you told me that before all the hours or trying to understand differential geometry
Actually i learned something new... thanks.

The problem comes when you try and “flatten” the sphere...without any distortion.

At what point have i tried to "flatten" the sphere?

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

The model has intrinsic curvature but noone says an orthonormal axis is the only way it can be viewed.
There are infinitely many more representations, and the only way to truly know the correct axis, if it even exists, is to look at the universe from the outside.
I think it all depends on how i want to look at reality.
I can see a ship disappearing over the horizon keel first and see it as evidence of curvature of the globe, or i can think of it as a flat earth with light curving upwards.
Neither is wrong, they're just different views on the same reality.

Personally I think people are way too attached to a preferred shape.

[WTF_Seriously]

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

My axis are the equator (circular) and the Greenwich greatcircle. What 'straight' path are you taking?
Straight is only defined the way you think it is in an orthonormal basis.
This drawing is in celestial coordinates (lat, lon, distances). Please don't treat it as (x,y,z) cartesian coords because that will indeed break everything.

Sorry, straight line path was not the best description for a globe model for 'straight line path' is actually an arc along the surface.

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.

So, taking my example if the distance from Sydney to LAX is 1 round, then the distance from Sydney to Santiago is roughly .94 rounds.  Regardless the coordinate system, this ratio must hold.

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.

[troolon]

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

Unfortunately I have no references to share. These people don't fancy half the internet mailing them for explanations.

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

For a funhouse mirror it's quite easy for an outside observer to see the shapes don't match with the owner.
Problem is we can't step outside of the universe to check. We have no absolute references.
If the universe were a simulation, how would you tell it's simulating a flat earth or a globe?
For someone inside the simulation, Australia will have the same size regardless. And you just can't know what the computer is calculating.

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

Of course it works in a flat representation of physics. It's just a representation, all predictions are the same. Can't be distinguished remember?

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

Speed of light is constant, in the flat-earth representation radio waves travel in curves. It's distance that's not what you think it is.

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

Same physics in both models. Indistinguishable.

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

never said the globe representation was wrong, quite the contrary.

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

Let's play a little game....
I'm going to design a universe. It's either going to be a globe with straight light, a flat earth with bendy light, or a rectangle earth with different bendy light.
I will place you in my little universe.
How will you and your ruler tell in what universe you are?
Remember for an inside observer all universes measure the same...

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

I agree physics works remarkably well. You're quoting comments on a different model I don't believe are applicable. It's the same physics. It's just a different representation. Bendy light stopped just before realizing this.

Where is Sigma Octantus?

I'll try to make an updated graph
But you seem to be missing the point: this flat earth is just a different representation of physics.
You're trying to debunk globe physics...

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

Theoretically yes. Practically that's not how the software is currently written, nor do i have immediate plans. If you desperately want to see it, make it yourself, i've explained the transformation, it's not hard:
- convert to celestial coordinates relative to the center of the earth
- draw latitude on a straight axis

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

My latest model with the planet of the solar system is verifiable with stellarium. The positions of the planets match the time.
I've actually been thinking about doing earth based views in software and it's just pointless. The way i would write it would be a bunch a coord transforms back and forth between celestial and cartesian coords, and they'd all cancel out. I'd rather save myself the trouble.
You still don't seem to be grasp my model is regular physics. It's just a different representation.
You're asking me to make drawings of regular physics.

But show how everything works when you move around the simulation on the surface. There's your homework.

I'm not taking homework from you.
Also i've deleted various paragraphs i deemed offensive.
I believe i've been very accommodating and polite with my replies to you. If you keep up the hostilities i will stop replying.

[JSS]

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

For a funhouse mirror it's quite easy for an outside observer to see the shapes don't match with the owner.
Problem is we can't step outside of the universe to check. We have no absolute references.
If the universe were a simulation, how would you tell it's simulating a flat earth or a globe?
For someone inside the simulation, Australia will have the same size regardless. And you just can't know what the computer is calculating.

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.

You might as well argue if whatever alien supercomputer that runs the universe uses binary or trinary logic circuits.  We can never know, and deciding it's one or the other can't produce any useful results.

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. 

[Clyde Frog]

It's not a theory. It's a model. It's quite literally just a coordinate transformation. It's amazing how much pushback there is over this.

[Tumeni]

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?

[Clyde Frog]

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.

[Iceman]

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

[Rog]

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?

I'll try to answer your latest post tonight. But i think in summary my reply will be:
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)
I believe you're taking a disc in celestial coords, and expecting it to behave like a disc in an orthonormal basis.
Intrinsic curvature doesn't see the coordinate axis directly, but it does notice distances, angles and tangent planes behaving relative to this axis.
You're basically taking my representation, changing the axis, distances, angles, ... and then exclaiming: look curvature is broken.

No, a disc will always have a curvature of zero and a sphere will always have a non-zero curvature. ]Changing coordinates doesn’t create curvature in an object where none physically exists.  You are conflating curvature of the object and a curvature of the coordinate system.  If you can make the curvature appear or disappear by changing coordinates, the curvature isn’t intrinsic to the space.  It is just an artifact of the coordinate system. And distances, angles and tangent planes won’t be represented the same. 

“Intrinsic” curvature means that it is a physical, immutable, property of the object.  Coordinate systems are just mathematical abstractions.  They don’t influence physical reality anymore than photo shop does. A coordinate transformation is a representation, but it isn’t always an accurate one.  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.

Since the curvature of the sphere does not vanish, it CANNOT BE LOCALLY ISOMETRICALLY MAPPED TO THE EUCLIDEAN PLANE

http://www.math.utah.edu/~treiberg/MappingtheEarthSlides.pdf

Luckily, though you don’t need to know or do any complicated math. If a flat circle had intrinsic curvature, you wouldn’t be able to parallel transport a vector around the circumference.

For a funhouse mirror it's quite easy for an outside observer to see the shapes don't match with the owner.
Problem is we can't step outside of the universe to check. We have no absolute references.

Yes, we do have an absolute reference.  See above.

At what point have i tried to "flatten" the sphere?

By trying to project a sphere onto a flat surface without any distortion.

The model has intrinsic curvature but noone says an orthonormal axis is the only way it can be viewed.

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.