The Flat Earth Society

Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: troolon on January 28, 2022, 10:45:57 PM

Title: Found a fully working flat earth model?
Post by: troolon on January 28, 2022, 10:45:57 PM
Hello,

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
Title: Re: Found a fully working flat earth model?
Post by: inquisitive on January 29, 2022, 08:43:28 PM
Hello,

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
And how about meaaured distances?
Title: Re: Found a fully working flat earth model?
Post by: jimster on January 30, 2022, 08:03:40 PM
Unless you explain the curved light and the distortion of distances (per Gauss' Remarkable Theorem), can you really call it a "working model"?

That's why FE says nobody knows the distances over the oceans (per Tom Bishop), and the Electromagnetic Acceleration FAQ page cites "unknown forces with unknown equations" to explain the bending light. It is why all the maps in the FAQ have wrong distances.

So basically, the FE position is that light bends in exactly the way it needs to for FE but we don't know how or why and we can't know the info necessary to determine the the shape of the earth by distance measurements. Interestingly, this does not prove FE, the earth could be any shape, including round, and we can never know.

Oh, won't someone please figure out the exact way the light is bending and figure out how to measure the distances over the oceans? Perhaps we can never know?
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 30, 2022, 10:52:34 PM
- Distances check out. The model has a special distance metric so Australia's width for example is correct.
- The equations for bended light rays are on my site. https://troolon.com
  In fact i can model anything globe-physics can on a flat earth it's fully equivalent.
-  Fermat's principle explains it i think. Light takes the fastest path, and because of the warped distance metric it checks out.

I wholeheartedly agree the earth can have any shape and we'll never know.
But it can now be shown flat is a possible shape and physics still works when it does.
Title: Re: Found a fully working flat earth model?
Post by: jimster on January 31, 2022, 02:56:36 AM
Can you make a graphic with the sun 93 million miles away and straight light path? Can you make a graphic of a globe map where all the distances and direction match observed? There are many many of these, I think you can. RE geometry works with straight light rays to explain day/night. Works without unexplained light bending.

The circumference of 80 degrees north or south is much less than at the equator. How does this project/transform/??? to a cylinder? The circumference at every latitude would be the same as the equator. To a disk? That's why Australia is always too large on the FAQ maps. This is possible on a sphere, how can 80 degrees north circumference be so much smaller than 80 degrees south on a disk FE? They should be equal. Per Gauss, you can't project, transform or ??? onto a cylinder or a disk and have both 80 degree circumferences be correct.

Your graphic of disk earth has Australia too big, just like the FAQ maps. Let's see your flat map with accurate size, distance, direction, and constant scale. You have math and graphics skills. Let's see it. Be the first!

Re day/night bendy light, do your equations account for the left/right bending? At noon on the equinox in Kampala, Uganda, the sun is rising in Rio De Janero. It appears to be directly east bearing 90 degrees, but on FE disk map, it looks like the bearing is actually about 45 degrees. And from Perth Australia at the same moment, looks like 270 degrees (directly west), but is actually 315 degrees.

Also interesting is that the light has no left right bending directly north and south. The amount of bend increases as you go south along the line of sunset until it gets to the far corner of the lighted area, where the discrepancy is maximum, then decreases until it is again 0 directly south.

So you need to add that to your model.

Interesting physical phenomenon, bends the opposite way on either side. As the sun moves around, this pattern moves with it. Apparently, the sun throws curveballs?

Now add the dome to your model. When it is sunset in Denver, in Salt Lake City it is still light blue over the entire dome. In St Louis, it is dark and there are stars over the entire sky, including the part of the dome beyond Salt Lake City. If someone in St Louis and someone in Salt Lake City look at the exact same spot over Denver, one sees light blue day sky, the other sees dark night with stars.

So let's see a model that accounts for left/right bending, and shows how someone can see either the entire dome of light blue at the same time as another person a few hundred miles east sees dark and stars, and has circumference of 80 degrees latitude both north and south much smaller than the equator on a flat disk, or cylinder, or anything other than a sphere.

You claim to have the ability to produce a FE map with accurate distance, size, direction, and constant scale. I await your post and will be astounded.










Title: Re: Found a fully working flat earth model?
Post by: jimster on January 31, 2022, 03:05:28 AM
My apologies, the graphics are small, but maybe you did take left/right into account.

You can't make Gauss Remarkable Theorem untrue by using polar coordinates. You still can't have 80 degrees north and south latitude both be smaller than the equator on a disk or a cylinder.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 07:31:40 AM
The map has a different distance metric. Distance is just a formula, it's up to you to choose a meaningful one.
Using the correct metric, the circumference of 80N and 80S is the same as the globe and so it's smaller than the equator.

I believe you might be taking an orthonormal straight ruler and be applying it to a non-orthonormal basis and saying everything's wrong.
Flat earth rulers are curved, the markings are not equally spaced, and you need a different shaped ruler depending on position and direction you're measuring. (not very practical but that's flat earth for ya)
If i were to take this flat-earth ruler to the sphere, it won't make any sense at all either.

Shapes are meaningless without an axis (see attached)
I'm not very familiar with gauss remarkable theorem, but i'm pretty sure it changes when your coord system changes.
IF i'm applying it correctly, the gauss number for my flat earth is 0 though.

If the universe has an orthonormal basis, then the earth is a sphere, however we'll never be able to tell in what coord system the universe is supposed to be seen.
When you look down at your feet, you could see a sphere with straight light,
Or you could see a flat earth with bendy light.
There's no measurement/observation that will ever be able to tell the difference. Both are possible, neither is provable.
Title: Re: Found a fully working flat earth model?
Post by: DuncanDoenitz on January 31, 2022, 10:45:33 AM
The map has a different distance metric. Distance is just a formula, it's up to you to choose a meaningful one.
Using the correct metric, the circumference of 80N and 80S is the same as the globe and so it's smaller than the equator.

I believe you might be taking an orthonormal straight ruler and be applying it to a non-orthonormal basis and saying everything's wrong.
Flat earth rulers are curved, the markings are not equally spaced, and you need a different shaped ruler depending on position and direction you're measuring. (not very practical but that's flat earth for ya)
If i were to take this flat-earth ruler to the sphere, it won't make any sense at all either.


If the Earth is flat, then any flat map (on paper or screen) is just a scale drawing, with a constant scale.  No need for bendy rulers. 

The scale is n, map distance is y, and actual distance is x.  If your formula is more complex than  x=ny, then the Earth isn't flat.
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on January 31, 2022, 10:50:16 AM
The map has a different distance metric. Distance is just a formula
What do you mean by that? This feels like a completely meaningless statement.
Distance is a physical thing which can be measured.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 12:01:23 PM
Distances in a non-orthonormal basis may not be measured with an orthogonal ruler.
You need a flat-earth-ruler to measure on a flat-earth map.

In maths: sqrt(xx+yy+zz) is not the correct formula. Apply that to Australia and you'll get gibberish.
You need to use a meaningful distance metric for the map/model you're using.
For the AE map, the conventional distance formula for coordinates expressed as (lat, long) does work as lat/long are preserved by the projection
Title: Re: Found a fully working flat earth model?
Post by: Kangaroony on January 31, 2022, 12:56:56 PM
Distances in a non-orthonormal basis may not be measured with an orthogonal ruler.
You need a flat-earth-ruler to measure on a flat-earth map.

In maths: sqrt(xx+yy+zz) is not the correct formula. Apply that to Australia and you'll get gibberish.
You need to use a meaningful distance metric for the map/model you're using.
For the AE map, the conventional distance formula for coordinates expressed as (lat, long) does work as lat/long are preserved by the projection

You're a bit early with this.  There's still 8 weeks to go before 1st April.
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on January 31, 2022, 01:58:41 PM
Distances in a non-orthonormal basis may not be measured with an orthogonal ruler.
You need a flat-earth-ruler to measure on a flat-earth map.
Sorry, but this is just gibberish.
If you drive along a straight road and the odometer tells you that you've gone 100 miles then, so long as it's accurate, you have travelled 100 miles.
That is true whether the earth is a sphere or flat. The only difference is on a globe that 100 miles will be part of a circle, on a flat earth it will be a straight line.
I'm ignoring the complications of terrain.
If the earth is flat then it should be possible to make an accurate flat earth map which depicts the shapes and sizes of land masses correctly and the distances between places.
Maps are flat, if the earth is flat too then the only issue should be scale. I have yet to see a flat earth map which does this and on which flight routes make sense.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 02:35:32 PM
To make an accurate map,
could you perhaps first tell me what coordinate system reality has?
I haven't been able to find the X-axis yet.  :)

But more seriously, i'm trying to make a model of reality, i'm not trying to make a map.
- Coordinate transformations can turn any shape into any other shape
- physics works with coordinate trnasformations
-> physics can be made to work on any shape universe (have a look at http://troolon.com for pictures)
-> There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape.

So have a look around you and try these two views: i'm standing on a globe and lightrays are straight,
or you could say: i'm standing on a flat plane, and light curves to exactly counteract the missing curve.
Your eyes wouldn't be able to tell the difference and there's no physical test to distinguish between the two views, it's just a matter of perception.

It's like the old question: Am i moving, or is the entire universe moving around me? It's just a matter of how you look at the world.

Also this result shouldn't be very surprising. The universe could already be a sphere, a simulation, have no shape (QM), be a restored backup from 5 minutes ago ... We will simply never know the shape of the planet. It can be flat, it can be a globe or even a velociraptor.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on January 31, 2022, 02:52:30 PM
Though your animations are nice, they aren't much different in concept than the sunlight animation already shown on the WIKI here.

Others are discussing the distance issue.  I'm curious how your model, appearing to be north pole centric, handles full daylight in the south in winter.  Take a look at the rightmost image here:

(https://i.imgur.com/V4g1PJP.png)

In winter, on a flat, north monopole, disc light forms a ring around the outside like depicted.  How does your model propose light does this?
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on January 31, 2022, 03:23:19 PM
To make an accurate map,
could you perhaps first tell me what coordinate system reality has?
I haven't been able to find the X-axis yet.  :)

But more seriously, i'm trying to make a model of reality, i'm not trying to make a map.
- Coordinate transformations can turn any shape into any other shape
- physics works with coordinate trnasformations
-> physics can be made to work on any shape universe (have a look at http://troolon.com for pictures)
-> There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape.

So have a look around you and try these two views: i'm standing on a globe and lightrays are straight,
or you could say: i'm standing on a flat plane, and light curves to exactly counteract the missing curve.
Your eyes wouldn't be able to tell the difference and there's no physical test to distinguish between the two views, it's just a matter of perception.

It's like the old question: Am i moving, or is the entire universe moving around me? It's just a matter of how you look at the world.

Also this result shouldn't be very surprising. The universe could already be a sphere, a simulation, have no shape (QM), be a restored backup from 5 minutes ago ... We will simply never know the shape of the planet. It can be flat, it can be a globe or even a velociraptor.

Again, it comes down to the point made about distance. If I drive between two points, the distance travelled is the distance travelled, whether it's flat, curved, rectangular or whatever. If it's flat, then it should be easy to lay out all the landmasses and say what the distances are between them, and indeed what size and shape they all are. How far is it, in your system, between the east and west coast of Australia, for example? How far is it from Australia to the southern tip of Africa or South America?
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on January 31, 2022, 03:55:31 PM
But more seriously, i'm trying to make a model of reality, i'm not trying to make a map.
A map is a model of reality, or part of one. As SteelyBob said, the distance between places can be measured.
If you know the distance between enough pairs of places you can determine whether it is possible to plot those on a plane. If it is not then one is forced to conclude that the earth cannot be flat.

Quote
There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape.
I'd suggest that astronauts going to space and observing the earth is a pretty good test, and things like GPS are good evidence for the shape of the earth.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 08:49:17 PM
Though your animations are nice, they aren't much different in concept than the sunlight animation already shown on the WIKI here.

Others are discussing the distance issue.  I'm curious how your model, appearing to be north pole centric, handles full daylight in the south in winter.  Take a look at the rightmost image here:

(https://i.imgur.com/V4g1PJP.png)

In winter, on a flat, north monopole, disc light forms a ring around the outside like depicted.  How does your model propose light does this?
Do you mean this one? The only one i haven't encountered yet is the last wobbly one but i suppose that's because i approximated the suns orbit with a circle? If you could give me lat/long of the sun for the last one, i'll whip an image out.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 08:58:37 PM
Again, it comes down to the point made about distance. If I drive between two points, the distance travelled is the distance travelled, whether it's flat, curved, rectangular or whatever. If it's flat, then it should be easy to lay out all the landmasses and say what the distances are between them, and indeed what size and shape they all are. How far is it, in your system, between the east and west coast of Australia, for example? How far is it from Australia to the southern tip of Africa or South America?

The AE map has a nonorthonormal basis. We can consider two cases:
For an observer existing within the coordinate system, ie a person in austraiia, the world and distances appear as in reality.
For an observer outside of the coordinate system, you should measure distances with a flat-earth ruler. (which is curved and has non-equal distance markings)
Taking an orthogonal ruler, to a flat-earth coordinate system produces invalid results. Just like taking my bend ruler to your globe would completely invalidate it.
Mathematically: you need to use the corect flat-earth distance metric that compensates for all the distortions.
This is why I asked you about the axis of reality. Globe is the correct shape if the axis is orthonormal. Flat is the correct shape if it's a peculiar other base. As we do not know in what base we're supposed the see the universe, we also do not know what shape the earth really has.


The way this model was constructed was by applying a coordinate transform to globe physics. Coord transforms change the shape of the model, but not the physics.
So if distances work on the globe, they will also work on the flat model, by design.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 09:03:47 PM
A map is a model of reality, or part of one. As SteelyBob said, the distance between places can be measured.
If you know the distance between enough pairs of places you can determine whether it is possible to plot those on a plane. If it is not then one is forced to conclude that the earth cannot be flat.
Please check the other post for distance.

Quote
I'd suggest that astronauts going to space and observing the earth is a pretty good test, and things like GPS are good evidence for the shape of the earth.
In this model light curves in exactly the correct way to make the flat earth look curved. This model is globe physics coord transformed. We removed the curve of the earth and contorted the entire universe to compensate for it. Light bends in exactly the correct way so you can't ever tell the difference.

This flat earth model is globe physics made to look like flat. It is indistinguishable from the globe model.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 09:06:32 PM
Hey,

This is how you make the any shaped earth:
- coordinate transformations can change the universe to any shape you like
- physics is unaffected by coordinate transformations
-> physics can be made to work on any shape world (see the animations i posted for example)
-> There's no test to distinguish the models, they're designed to be identical
-> It's impossible to ever know the shape of the planet
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on January 31, 2022, 09:24:57 PM
Though your animations are nice, they aren't much different in concept than the sunlight animation already shown on the WIKI here.

Others are discussing the distance issue.  I'm curious how your model, appearing to be north pole centric, handles full daylight in the south in winter.  Take a look at the rightmost image here:

(https://i.imgur.com/V4g1PJP.png)

In winter, on a flat, north monopole, disc light forms a ring around the outside like depicted.  How does your model propose light does this?
Do you mean this one? The only one i haven't encountered yet is the last wobbly one but i suppose that's because i approximated the suns orbit with a circle? If you could give me lat/long of the sun for the last one, i'll whip an image out.

Hard to tell on your image what areas are supposed to be in daylight and what are not.  Also, it appears the sun is positioned directly over the north pole which is not what happens.  As for what I was showing in the right hand image, that would be the sun at any longitude and a latitude of 23. south deg. south.  At that time, on a north pole centric disc there would be a 'spotlight' darkness surrounded by a ring of light lighting the southern regions in 24 hour daylight.
Title: Re: Found a fully working flat earth model?
Post by: troolon on January 31, 2022, 10:59:32 PM
Yeah, perspective made the sun look wrong, I've taken a top down photo and placed the sun over Africa so you can check the latitude to be -23.
Inside the circle made by the yellow dots it's dark. Outside that circle it's day. The white lines are rays of sun, traveling from the sun to the places where it's sunset/rise.
What i believe makes my work different is that i have actual equations for all optics, and i can correctly explain anything else.
As this model is based on a coordinate transformation it is equally powerful as globe physics.
That is because this IS the globe model. Coord transformations preserve physics, and only change shape.
There is no test to differentiate between the two models.
(http://troolon.com/wp-content/uploads/2022/01/sun_antarctica.png)

Also it can be proven that flat is a possible shape of the earth, but the true shape of the earth can never be proven.
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on January 31, 2022, 11:41:18 PM

The AE map has a nonorthonormal basis. We can consider two cases:
For an observer existing within the coordinate system, ie a person in austraiia, the world and distances appear as in reality.
For an observer outside of the coordinate system, you should measure distances with a flat-earth ruler. (which is curved and has non-equal distance markings)
Taking an orthogonal ruler, to a flat-earth coordinate system produces invalid results. Just like taking my bend ruler to your globe would completely invalidate it.
Mathematically: you need to use the corect flat-earth distance metric that compensates for all the distortions.
This is why I asked you about the axis of reality. Globe is the correct shape if the axis is orthonormal. Flat is the correct shape if it's a peculiar other base. As we do not know in what base we're supposed the see the universe, we also do not know what shape the earth really has.


The way this model was constructed was by applying a coordinate transform to globe physics. Coord transforms change the shape of the model, but not the physics.
So if distances work on the globe, they will also work on the flat model, by design.

Forget maps for second. Get in a car, and drive from one coast of Australia to the other. How far does the odo say you've travelled?
Title: Re: Found a fully working flat earth model?
Post by: stevecanuck on February 01, 2022, 12:13:27 AM

Forget maps for second. Get in a car, and drive from one coast of Australia to the other. How far does the odo say you've travelled?

Or ask anyone in charge of buying materials for building such roads, laying pipe, stringing phone lines, etc. They base the amount the asphalt, pipe, and line on known mileage which just happens to comply with RE distances.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 12:27:19 AM
 ;D You just couldn't have picked a a worse example to prove RE  ;D
Did you know the haversine formula works on AE maps unmodified?

spherical distances are calculated using the haversine formula, which takes as input lat/long
lat/long is an invariant for the AE projection and so the globe formula will still work for the AE map unmodified.

The distance is the same number the globe physics model says it is and thus the same as my flat model says. (because the models are equivalent)
Title: Re: Found a fully working flat earth model?
Post by: stack on February 01, 2022, 05:53:12 AM
;D You just couldn't have picked a a worse example to prove RE  ;D
Did you know the haversine formula works on AE maps unmodified?

spherical distances are calculated using the haversine formula, which takes as input lat/long
lat/long is an invariant for the AE projection and so the globe formula will still work for the AE map unmodified.

The distance is the same number the globe physics model says it is and thus the same as my flat model says. (because the models are equivalent)

True, Haversine is independent of projections. But where I'm kinda hung up is still on a few things mentioned by others and maybe some not mentioned:

- Isn't it a little strange that the AE projection you're basing this all on is just that, a projection...of a globe?
- Isn't Haversine based on spherical geometry, specifically a great circle on a ball?
- In geodesy, Haversine may be used for the quick and dirty shorter calculations, but predominantly Vincenty is used because it's more accurate, based upon an ellipsoid rather than a sphere. That would seem to alter the look of the AE projection to a degree
- You mentioned something about a depiction of reality and not a map. I get that, but what's a "flat earth ruler" and where can I find one because it doesn't seem to exist in said reality

All of this gets wonky pretty fast when you start mixing projections with reality (navigation) with useable maps (navigation). For instance, the AE projection is a projection from a globe. As mentioned before, because projections are required for a globe onto a flat surface, distortions arise. For a flat earth depiction on a flat surface, no distortion is required, it's one-to-one. And it would actually resemble reality. And as I think AATW mentioned, I too have yet to see a non-distorted depiction of a flat earth on a flat surface.
Then there's the reality of navigation. How does that work exactly? Ex., QANTAS JND>SYD flight, the distance may be able to be calculated correctly on the AE, but from a reality/visual perspective, the route makes zero sense.

And of course, little things like how have people been able to circumnavigate Antarctica when it's a ring walling in the earth.

Then there's the whole bendy light rabbit hole which, well, probably needs a whole separate discussion around that.

Curious about your thoughts and thanks for pulling all this stuff together, makes for an interesting examination.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on February 01, 2022, 09:27:55 AM
Isn't it a little strange that the AE projection you're basing this all on is just that, a projection...of a globe?
This continues to be false, no matter how many times you say it. You already know this, so please stop trying to propagate this lie.
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on February 01, 2022, 10:34:15 AM
Isn't it a little strange that the AE projection you're basing this all on is just that, a projection...of a globe?
This continues to be false, no matter how many times you say it. You already know this, so please stop trying to propagate this lie.

If it's not a projection of a globe, then what is it? Where did the shapes, sizes and positions of the various land masses come from, if not the globe? The lat/long relationship of each significant feature appears to have been preserved, unless I'm mistaken - do correct me if I'm wrong.
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on February 01, 2022, 11:24:46 AM
;D You just couldn't have picked a a worse example to prove RE  ;D
Did you know the haversine formula works on AE maps unmodified?

spherical distances are calculated using the haversine formula, which takes as input lat/long
lat/long is an invariant for the AE projection and so the globe formula will still work for the AE map unmodified.

The distance is the same number the globe physics model says it is and thus the same as my flat model says. (because the models are equivalent)

But the output of the haversine formula is the great circle distance between two points. If the earth was flat, then that wouldn't be the distance between two points, it would be some other number.

Again, if I get in a car and drive from east to west coast of Australia, what would my odo say the distance was? I would suggest it would say roughly 2500 miles or thereabouts. If the world was indeed flat, and laid out as per the AE / monopole map as shown in the wiki and your proposal, then the distances would be roughly 2.4 times greater, based on a latitude of -30 degrees, so around 6000 miles.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on February 01, 2022, 01:06:23 PM
If it's not a projection of a globe, then what is it? Where did the shapes, sizes and positions of the various land masses come from, if not the globe? The lat/long relationship of each significant feature appears to have been preserved, unless I'm mistaken - do correct me if I'm wrong.
The most commonly shared image is, if I remember correctly, a re-projection of a Mercator map, which is where the lat-lon lines were introduced. However, the map you refer to as "the AE map" predates RE by centuries.
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on February 01, 2022, 01:46:49 PM
The most commonly shared image is, if I remember correctly, a re-projection of a Mercator map, which is where the lat-lon lines were introduced. However, the map you refer to as "the AE map" predates RE by centuries.


In response to:
Quote
Isn't it a little strange that the AE projection you're basing this all on is just that, a projection...of a globe?

you said:
Quote
This continues to be false, no matter how many times you say it

So what exactly are you saying? That the AE map isn’t a projection? That the basis of the OP’s work isn’t a projection? The existence, or not, of some vaguely referenced historic map has nothing to do with whether or not the AE map, and indeed the same map as presented in the wiki, is a projection.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 01, 2022, 03:00:57 PM
Yeah, perspective made the sun look wrong, I've taken a top down photo and placed the sun over Africa so you can check the latitude to be -23.
Inside the circle made by the yellow dots it's dark. Outside that circle it's day. The white lines are rays of sun, traveling from the sun to the places where it's sunset/rise.
What i believe makes my work different is that i have actual equations for all optics, and i can correctly explain anything else.
As this model is based on a coordinate transformation it is equally powerful as globe physics.
That is because this IS the globe model. Coord transformations preserve physics, and only change shape.
There is no test to differentiate between the two models.
(http://troolon.com/wp-content/uploads/2022/01/sun_antarctica.png)

Also it can be proven that flat is a possible shape of the earth, but the true shape of the earth can never be proven.

Interesting.  Curious.  You have equations for the optics that explain how light travels different distances and curves differently simply based off of the latitude of the sun?

Also, can you generate a similar image of what the light pattern looks like when the sun's on the equator
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 01, 2022, 03:05:00 PM
the true shape of the earth can never be proven.

Why not?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 03:39:17 PM
Consider this thought experiment: that shows the earth can have any possible shape and we'll never be able to prove the correct one.

- coordinate transformations can turn any shape into any other shape (math cares about numbers not drawings)
- coordinate transformations don't break physics (celestial coords, cartesian, polar, ...  physics uses all of these all the time and doesn't break)
-> physics can be made to work on any shape universe
-> There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape.
-> we can't ever test the shape of the earth. It can be flat, it can be a globe or a duck, impossible to tell from within the universe.

So have a look around you and try these two views: i'm standing on a globe and lightrays are straight,
or you could say: i'm standing on a flat plane, and light curves to exactly counteract the missing curve.
Your eyes wouldn't be able to tell the difference and there's no physical test to distinguish between the two views, it's just a matter of perception.

It's like the question: Am i moving, or is the entire universe moving around me? It's just a matter of how you look at the world.

Also this result shouldn't be very surprising. The universe could already be a sphere, a simulation, have no shape (QM), be a restored backup from 5 minutes ago ... We will simply never know the shape of the planet. It can be flat, it can be a globe or even a velociraptor.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on February 01, 2022, 04:05:45 PM
So what exactly are you saying?
Exactly what I said already, nothing more, nothing less. If you're trying to find a second meaning in it, you're doing it wrong.

That the AE map isn’t a projection?
I object to the naming of "the AE map" - that implies that it was originally produced through the azimuthal equidistant projection, which it wasn't.

That the basis of the OP’s work isn’t a projection?
Indeed.

The existence, or not, of some vaguely referenced historic map has nothing to do with whether or not the AE map, and indeed the same map as presented in the wiki, is a projection.
This is incorrect. While you can recreate reasonably good images that closely resemble the FE map through the use of the AE projection, it is not how the map came to be. This is essential basic knowledge without which you should not be posting here. Stack does have this knowledge, and is deliberately spreading misinformation - there was no need for me to go into much detail in order to plainly point out to him that he'll be stopping immediately.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 01, 2022, 04:10:07 PM
Consider this thought experiment: that shows the earth can have any possible shape and we'll never be able to prove the correct one.

- coordinate transformations can turn any shape into any other shape (math cares about numbers not drawings)
- coordinate transformations don't break physics (celestial coords, cartesian, polar, ...  physics uses all of these all the time and doesn't break)
-> physics can be made to work on any shape universe
-> There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape.
-> we can't ever test the shape of the earth. It can be flat, it can be a globe or a duck, impossible to tell from within the universe.

So have a look around you and try these two views: i'm standing on a globe and lightrays are straight,
or you could say: i'm standing on a flat plane, and light curves to exactly counteract the missing curve.
Your eyes wouldn't be able to tell the difference and there's no physical test to distinguish between the two views, it's just a matter of perception.

It's like the question: Am i moving, or is the entire universe moving around me? It's just a matter of how you look at the world.

Also this result shouldn't be very surprising. The universe could already be a sphere, a simulation, have no shape (QM), be a restored backup from 5 minutes ago ... We will simply never know the shape of the planet. It can be flat, it can be a globe or even a velociraptor.

Consider this thought experiment.  I'm in the ISS and I circle the planet every 90-ish minutes.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 04:49:39 PM
Dear Sirs,

My model IS globe physics expressed in celestial coordinates (lat, long, distance).

- The ISS works in the model because it works in globe physics
- distances work in the model because they work in globe physics
- Vincenti works in the model because it works in globe physics
- light rays work in the model because it works in globe physics
- ...
You are trying to debunk globe physics in celestial coordinates.

It is very surprising, but physics works regardless of shape. Math doesn't care how it's drawn!
And there is perhaps an important detail to highlight: You are looking at the model from outside the universe like a God. Gods can see the true shape of the universe. For an observer inside the universe, they all behave the same. From inside the universe it's impossible to tell them apart as all these models physics ARE THE SAME.

(http://troolon.com/wp-content/uploads/2022/01/flat_seasons.gif)
(http://troolon.com/wp-content/uploads/2022/01/capped_cube_seasons.gif)
(http://troolon.com/wp-content/uploads/2022/01/snowglobe.gif)
(http://troolon.com/wp-content/uploads/2022/01/illuminati_seasons.gif)

The implications of this is that flat earth has been proven possible. As has any other shape universe. We will never be able to tell.
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on February 01, 2022, 04:51:20 PM

That the basis of the OP’s work isn’t a projection?
Indeed.

So the OP is wrong about his/her own work? From the linked site:

Quote
There exists a coordinate transformation that transforms 3D cartesian space into an infinitely high cylinder, whereby spheres around the origin are transformed into their azimuthal equidistant projection.

The entire concept is based around an AE projection.

Quote
While you can recreate reasonably good images that closely resemble the FE map through the use of the AE projection, it is not how the map came to be. This is essential basic knowledge without which you should not be posting here. Stack does have this knowledge, and is deliberately spreading misinformation - there was no need for me to go into much detail in order to plainly point out to him that he'll be stopping immediately.

Leaving aside the OP and his work, could you enlighten me and tell me how the FE map came to be then? The wiki appears to be lacking in this department, unless I'm missing something? https://wiki.tfes.org/Flat_Earth_Maps (https://wiki.tfes.org/Flat_Earth_Maps)

What process, if not AE projection, was used to create this map?

(https://wiki.tfes.org/images/4/43/Map.png)

Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on February 01, 2022, 04:54:32 PM
The implications of this is that flat earth has been proven possible. As has any other shape universe. We will never be able to tell.
This continues to be untrue no matter how many times you say it.
Distance is not a mathematical concept, it's a physical thing. You can travel between places and measure the physical distance between them.
If you do that with enough pairs of places you can then determine whether it is possible to plot those places and the known distances between them on a flat plane. If you can't then the earth cannot be flat.
Title: Re: Found a fully working flat earth model?
Post by: SteelyBob on February 01, 2022, 04:59:04 PM
Dear Sirs,

My model IS globe physics expressed in celestial coordinates (lat, long, distance).

- The ISS works in the model because it works in globe physics
- distances work in the model because they work in globe physics
- Vincenti works in the model because it works in globe physics
- light rays work in the model because it works in globe physics
- ...
You are trying to debunk globe physics in celestial coordinates.

It is very surprising, but physics works regardless of shape. Math doesn't care how it's drawn!
And there is perhaps an important detail to highlight: You are looking at the model from outside the universe like a God. Gods can see the true shape of the universe. For an observer inside the universe, they all behave the same. From inside the universe it's impossible to tell them apart as all these models physics ARE THE SAME.

The implications of this is that flat earth has been proven possible. As has any other shape universe. We will never be able to tell.

I note you seem to be evading my point about actual distances experienced. You can't just wave it away.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on February 01, 2022, 05:16:34 PM
So the OP is wrong about his/her own work?
No.

Stop trying to derail this thread. Final warning.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 05:17:59 PM
Physics is a model.
It's a good model if it produces the same results as we measure in reality.
The distance metric in my model gives the exact same answer as the globe model as the two are the same.
Shapes and distances only make sense compared to a basis.

(http://troolon.com/wp-content/uploads/2022/02/what_shapes_are_these.png)
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 07:01:32 PM
Please find the sun_at_the_equator image hereby.

If you prefer, it's relatively easy to make these graphs yourself:
- generate the scene as the globe model describes (ie, an earth, a sun and straight lightrays from the sun to the earth)
- express every element in celestial coordinates (latitude, longitude, distance from the center of the earth)
- now draw (lat/long) as an AE projection (this creates a disc) and insert this disc at height `distance` in the cylinder.
Doing this for all elements in the scene will create the pictures i've made.

Or if you like hard math, transform the line-equation this way, and you'll have the equation for lightrays in the flat-earth universe.
(http://troolon.com/wp-content/uploads/2022/02/sunequator.png)
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 01, 2022, 07:09:21 PM
Please find the sun_at_the_equator image hereby.

If you prefer, it's relatively easy to make these graphs yourself:
- generate the scene as the globe model describes (ie, an earth, a sun and straight lightrays from the sun to the earth)
- express every element in celestial coordinates (latitude, longitude, distance from the center of the earth)
- now draw (lat/long) as an AE projection (this creates a disc) and insert this disc at height `distance` in the cylinder.
Doing this for all elements in the scene will create the pictures i've made.

Or if you like hard math, transform the line-equation this way, and you'll have the equation for lightrays in the flat-earth universe.
(http://troolon.com/wp-content/uploads/2022/02/sunequator.png)

You posted another animation which showed this.  Thanks for the still shot though.

Still interested to know the maths and, more importantly, the mechanism which causes light to travel different distances and in different paths from a single light source in your model.

Title: Re: Found a fully working flat earth model?
Post by: Dr David Thork on February 01, 2022, 07:28:01 PM
Please find the sun_at_the_equator image hereby.

If you prefer, it's relatively easy to make these graphs yourself:
- generate the scene as the globe model describes (ie, an earth, a sun and straight lightrays from the sun to the earth)
- express every element in celestial coordinates (latitude, longitude, distance from the center of the earth)
- now draw (lat/long) as an AE projection (this creates a disc) and insert this disc at height `distance` in the cylinder.
Doing this for all elements in the scene will create the pictures i've made.

Or if you like hard math, transform the line-equation this way, and you'll have the equation for lightrays in the flat-earth universe.
(http://troolon.com/wp-content/uploads/2022/02/sunequator.png)

You would only expect those sunlight patterns on 20th March and 23rd September.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 07:32:55 PM
The mechanism by which light travels is the same as on the globe.
If you can tell me why light travels straight in the globe model, i'll give you the same answer for a flat earth :)
It just does, it matches observations.

for more maths you can check the site. https://troolon.com.
The computer code that generates the images has these functions:
      xyz_to_celest(), celest_to_ae() and ae_to_screen()
We then chain them together as: ae_to_screen(celest_to_ae(xyz_to_celest(coordinate)))
Or coloquially: express in celestial coords (lat, long, distance), draw the celestial coords as AE cylinder, plot on screen....

This model will work, and all of physics will work, but the maths will be a lot harder than the globe.
The only use of this model is to show that it's possible the earth is flat, (or square, or a globe or duckshaped),
However the true shape of the planet can never be deduced from within this. Only an outside observer can tell us if we truly live on a duck :)
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 01, 2022, 07:50:24 PM
The mechanism by which light travels is the same as on the globe.

Well, not exactly.  Light travels the same distance in all directions from a single light source.   In your model, it does not.  In fact, not only does it not travel equal distance in all directions but the distance that it is able to travel in a given direction changes with position of the sun.  Why is that?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 08:06:53 PM
The mechanism by which light travels is the same as on the globe.

Well, not exactly.  Light travels the same distance in all directions from a single light source.   In your model, it does not.  In fact, not only does it not travel equal distance in all directions but the distance that it is able to travel in a given direction changes with position of the sun.  Why is that?
The distance metric is totally whacked and the speed of light is the same. So the light looks also pretty wacked.
But again it's a coordinate transformation. You're trying to debunk globe physics expressed in spherical coordinates...
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on February 01, 2022, 10:10:21 PM
(http://troolon.com/wp-content/uploads/2022/02/what_shapes_are_these.png)
This is nonsense.
A square has certain properties - 4 equal length sides, 90 degree corners. These are physical things which can be measured.
If the four sides are different lengths - compared to each other - then it's not a square.
If the angles are not all 90 degrees - which is a definition, and the angles can be compared against each other - then it's not a square.
You can define these things in different ways, but that doesn't mean the different things equivalent.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 01, 2022, 10:20:28 PM
The mechanism by which light travels is the same as on the globe.

Well, not exactly.  Light travels the same distance in all directions from a single light source.   In your model, it does not.  In fact, not only does it not travel equal distance in all directions but the distance that it is able to travel in a given direction changes with position of the sun.  Why is that?
The distance metric is totally whacked and the speed of light is the same. So the light looks also pretty wacked.
But again it's a coordinate transformation. You're trying to debunk globe physics expressed in spherical coordinates...

I'm not trying to debunk anything.  I'm talking about this animation.

(http://troolon.com/wp-content/uploads/2022/01/snowglobe.gif)

It shows light changing how far it travels as well as how much it curves.  As others have said, distance is distance whether you're measuring it on a flat surface or curved.  Your 'distance metric' is nonsense.  A point source of light will emit light radially the same distance in all directions if not obstructed.  Your model doesn't show that to be the case.

Something else I noticed about your animation.  It appears the sun is changing elevation as it changes the shape of the pattern of light on the disc.  Is this the case in your model?
Title: Re: Found a fully working flat earth model?
Post by: GoldCashew on February 01, 2022, 10:37:19 PM
Please find the sun_at_the_equator image hereby.

If you prefer, it's relatively easy to make these graphs yourself:
- generate the scene as the globe model describes (ie, an earth, a sun and straight lightrays from the sun to the earth)
- express every element in celestial coordinates (latitude, longitude, distance from the center of the earth)
- now draw (lat/long) as an AE projection (this creates a disc) and insert this disc at height `distance` in the cylinder.
Doing this for all elements in the scene will create the pictures i've made.

Or if you like hard math, transform the line-equation this way, and you'll have the equation for lightrays in the flat-earth universe.
(http://troolon.com/wp-content/uploads/2022/02/sunequator.png)


In your above image at the left, the Sun's rays project light in only a "downward" direction from a single point.

Would not a spherical Sun emit rays in all directions though? If this be the case, than wouldn't you have rays projecting sideways  from the Sun and then bending downwards (due to bendy light) towards the flat Earth surface that you show as not lit?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 10:52:31 PM
This is nonsense.
A square has certain properties - 4 equal length sides, 90 degree corners. These are physical things which can be measured.
If the four sides are different lengths - compared to each other - then it's not a square.
If the angles are not all 90 degrees - which is a definition, and the angles can be compared against each other - then it's not a square.
You can define these things in different ways, but that doesn't mean the different things equivalent.
Math doesn't care how you draw things. It can't see. For math all these shapes are rectangles.
For an observer inside the coordinate system, all these shapes are also rectangles.
For an observer outside of the system, none of these are a rectangle.
This is the way we construct our universes. From the inside they're identical to globe physics, they're only different from the outside.
But as they're all identical on the inside, we can never know the true shape of the universe.
Title: Re: Found a fully working flat earth model?
Post by: scomato on February 01, 2022, 11:23:45 PM
I have one question about what people see on the ground, vs. what the map describes. In the animated diagram below you have light from the Sun warping every which way, stretching and contracting as needed to illuminate the Earth as we observe it. Wouldn't the sun appear to be wildly contorted in the sky, as the light from one part of the Sun will go on a roller coaster ride around Antarctica, while another will fall straight down? The shape of the sun in the sky should appear to alternate between an elliptical and crescent-moon shape with the changing seasons.

(http://troolon.com/wp-content/uploads/2022/01/snowglobe.gif)

Yet what we actually observe when we look at the Sun is a perfect sphere of unchanging size disappearing over the horizon. You can see dark spots on the Sun, which refutes the idea that sunlight takes different diverging paths to arrive at their special destinations, depending on where it is emitted from. From the observer's perspective, the light from the Sun has taken a straight-line path between its surface and your eyes. Regardless of whether the light was emitted from the top of the Sun or the bottom, straight-line path.

https://www.youtube.com/watch?v=pxWY7Tiy-UU

So the only possible explanation I can see is that spacetime itself is warped in between the Sun and the ground observer, creating the illusion that the light travelled in a straight line, configured precisely to simulate the appearance of a massive Sun-like object in distant space. But this introduces more problems than it answers - warping space to that degree would require a phenomenal source of mass or energy - that somehow only acts upon sunlight, without interfering with other forms of matter or energy on Earth.
Title: Re: Found a fully working flat earth model?
Post by: DuncanDoenitz on February 01, 2022, 11:46:29 PM
This is nonsense.
A square has certain properties - 4 equal length sides, 90 degree corners. These are physical things which can be measured.
If the four sides are different lengths - compared to each other - then it's not a square.
If the angles are not all 90 degrees - which is a definition, and the angles can be compared against each other - then it's not a square.
You can define these things in different ways, but that doesn't mean the different things equivalent.
Math doesn't care how you draw things. It can't see. For math all these shapes are rectangles.
For an observer inside the coordinate system, all these shapes are also rectangles.
For an observer outside of the system, none of these are a rectangle.
This is the way we construct our universes. From the inside they're identical to globe physics, they're only different from the outside.
But as they're all identical on the inside, we can never know the true shape of the universe.
I assume you are not a carpet fitter by trade. You would experience some very surprised customers. 

If that sounds facetious, it isn't meant to be.  Forget the complex maths (and I'm not convinced that you're not trolling), but if you can't get your head around the practicality of measuring something you are physically experiencing (be it with a straight ruler, odometer or whatever), and transferring those dimensions onto a model with which you are interacting (be that an engineering drawing, map or mental concept), I don't think you are really in a position to postulate the shape of the cosmos. 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 11:49:38 PM
It shows light changing how far it travels as well as how much it curves.  As others have said, distance is distance whether you're measuring it on a flat surface or curved.  Your 'distance metric' is nonsense.  A point source of light will emit light radially the same distance in all directions if not obstructed.  Your model doesn't show that to be the case.

Something else I noticed about your animation.  It appears the sun is changing elevation as it changes the shape of the pattern of light on the disc.  Is this the case in your model?
Imagine i have 2D cartesian coordinates (0,0) and (4, 4).  Then the distance formula would be sqrt((x1-x2)² + (y1-y2)²
In polar coordinates these coordinates would be (0°, 0) en (45°, 4).  If i plug these numbers into the distance formula, i get total gibberish. When you do a coord transform, you must update all formulas. That is what i mean with a new distance metric. It's the old one with compensation for the coord transform.

The animations show both day and night  and seasons. In this animation a year is 3 days long (just to keep the filesize reasonable)
So in all the animations, the sun is both rotating and moving from -23° to +23°.
Because of the dome, yes the sun seems to move up and down, but only for an outside observer.
For someone inside the universe/coordinate system, the height change is not measurable. From the inside this still looks like the reality globe physics describes.

But to understand the grand picture, all you should understand is that
- Physics supports coordinate transforms. Physics does not break if i make my measurements in celestial coordinates.
- coordinate transformations can change any shape into any other shape
=> physics works in any shape universe
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 01, 2022, 11:59:38 PM
I have one question about what people see on the ground, vs. what the map describes. In the animated diagram below you have light from the Sun warping every which way, stretching and contracting as needed to illuminate the Earth as we observe it. Wouldn't the sun appear to be wildly contorted in the sky, as the light from one part of the Sun will go on a roller coaster ride around Antarctica, while another will fall straight down? The shape of the sun in the sky should appear to alternate between an elliptical and crescent-moon shape with the changing seasons.
When you draw a sun on an AE-map (or run a sphere through an AE-projection) it will become distorted. The sun is no longer a sphere.
The light curving compensates exactly for this distortion and an observer will see it as perfectly circular again.
This is because the flat-earth universe is nothing but a coordinate transformation of the old one.
If i make the universe 2 times bigger (you and your ruler included), would you notice?
From inside the coordinate system you don't notice the change, everything appears as before. It's only for an outside observer that the universe appears flat.
From inside the universe looks exactly identical to the globe physics model.


Title: Re: Found a fully working flat earth model?
Post by: GoldCashew on February 02, 2022, 12:13:49 AM
Please find the sun_at_the_equator image hereby.

If you prefer, it's relatively easy to make these graphs yourself:
- generate the scene as the globe model describes (ie, an earth, a sun and straight lightrays from the sun to the earth)
- express every element in celestial coordinates (latitude, longitude, distance from the center of the earth)
- now draw (lat/long) as an AE projection (this creates a disc) and insert this disc at height `distance` in the cylinder.
Doing this for all elements in the scene will create the pictures i've made.

Or if you like hard math, transform the line-equation this way, and you'll have the equation for lightrays in the flat-earth universe.
(http://troolon.com/wp-content/uploads/2022/02/sunequator.png)


In your above image at the left, the Sun's rays project light in only a "downward" direction from a single point.

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


Hi. I think you may have skipped over the above inquiry.

Was curious to get your take.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 12:47:15 AM
I assume you are not a carpet fitter by trade. You would experience some very surprised customers. 

If that sounds facetious, it isn't meant to be.  Forget the complex maths (and I'm not convinced that you're not trolling), but if you can't get your head around the practicality of measuring something you are physically experiencing (be it with a straight ruler, odometer or whatever), and transferring those dimensions onto a model with which you are interacting (be that an engineering drawing, map or mental concept), I don't think you are really in a position to postulate the shape of the cosmos.
Given the time i've spent on programming the animations, i would be a very determined troll :)
Also i've had this work verified by at least 10 physicists by now. It is correct, i just find it very difficult to explain.
I wholeheartedly agree that this model is not practical. The only uses found so far are:
- to show it is possible to construct a flat earth model
- From this it can be deduced a flat earth is possible, but so is a globe, a pyramid or any shape universe, however it will be impossible for us to ever test, as all models will always predict all the same values for every test
- maybe there could be some value in visualization. I would love to be able to make space tangible and hold it in my hand.

Maybe i should update the title of the post to "Flat earth possible, but true shape of the universe unprovable"
As i spent a lot of time on the animations, i do tend to get bogged down into the details which i can imagine would be confusing.

Maybe I can explain better in 2D:
Imagine a cartesian coordinate grid with a line in it, terminated by points p1 and p2.
Also imagine you have some formulas that work on lines (say a length-formula)
- Now, we will express p1 and p2 in polar coordinates. Mathematically there's nothing wrong, but all our formulas break.
  We can fix our formulas by reverting back to cartesian coords.  so length_in_polar(polar1, polar2) = length_in_cartesian(polar_to_cart(polar1), polar_to_cart(polar2))
  The details don't really matter, just that all formulas can be made to work when we switch to polar coords.
  All right we now have done a coordinate transform, and everything still works.
- Now we will draw these polar coords funnily. An angle is a value between [0° and 360°].
  What if we were to draw our coordinate (angle, dist) onto a orthonormal X/Y grid.  Ie X ranging from [0°, 360°] and Y just the distance.
  Suddenly our line, would become an arc.
  It might seem crazy to draw angles linearly on an X-axis, but remember that in radians, the angle represents also the arclength of a unitcircle. It's also a length.
  However this is just a representation. Mathematically nothing changed. The polar coordinates are still the same we just drew them differently.
Conclusion: We did a little magic trick whereby we changed a line into an arc, while all maths and formulas still treat it as a line.

That's what i'm doing in 3D and with all of physics:
- We take any point in x,y,z-coordinates and transform to celestial coordinates. (latitude, longitude, distance)
  We must also update all formulas in physics to support this change. For example the distance formula needs to be updated as it works on cartesian instead of celestial coords.
  This is not impossible, in fact physicists use celestial coords and coordinate transformation all the time.
- Now we draw our celestial coords funnily. We do the same trick as before where we draw latitude (an angle) on a straight axis instead of as an angle. This effectively creates and AE-projection like effect, and the result is the flat earth you've seen.

So we've taken the globe, and all of physics and transformed it into a flat earth, with all the physics intact.
That's it. Hope it makes more sense like this.

Philosophical implications:
The trick above generalizes to all coord transforms. You can practically turn any shape into any other shape.
And physics, well they use coord transforms all the time.
So all of physics basically works on any shape universe.

From this we can conclude that the universe can have pretty much every possible shape
As all the models always produce the same answer for every question, we'll never be able to test which model is correct and thus what shape the universe has.
Earth could be a globe, or flat or ..... Any shape is possible, none is provable. At least not from within.
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on February 02, 2022, 09:16:10 AM
Math doesn't care how you draw things. It can't see. For math all these shapes are rectangles.
But there is an underlying physical reality. One which can be observed and measured.
If I go 1m in a direction on a flat plane, turn 90 degrees to my right and go another 1m and repeat that 2 more times I will be back where I started and will have traced out a square with side length 1m.
You can change your units, redefine 1m as 1km. OK, so now you have a square with side length 1km.
You've redefined the units but that doesn't mean the square is now 1000 times bigger on each side.
Or you could change the units so that units in one direction are twice as long as the units in the other. OK, so now you have a square 1m by 2m. It's still a square, it hasn't changed shape.
You could redefine the angle units so now we have a square which has 45 degree internal angles. Again, that doesn't change the shape of the object.
So sure, you can do all kinds of transformations but there is an underlying physical reality which can be observed and measured.
Being "inside" a coordinate system is a meaningless concept.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 11:36:32 AM
There is indeed but one reality.
And if i measure/observe something in reality, and i check with the globe model, the model will say everything checks out.
And when i check in the flat model, everything will also check out.
So both models explain/predict reality correctly.

And what does reality really look like? We don't know. It could look like one of the models, or it could be a simulation, or it could not even have a shape.
There's no way to tell.

Being "inside" a coordinate system is a meaningless concept.
You are quite right, and yet you are assuming you're in an orthonormal one because that's the only basis where earth looks like a sphere.
Ultimately a model is just a representation of reality and you can represent it in infinitely many ways.
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on February 02, 2022, 12:24:53 PM
And what does reality really look like? We don't know.
Yes we do. It has been measured and observed and there's only one shape which makes sense with those observations and measurements.
All you've done is show that you can map the reality on to different shapes.
People have been doing that for centuries with different map projections.
This is a good site which lets you play around with one such projection and see how the size of countries varies depending on where you place them.

https://thetruesize.com/

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

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Ultimately a model is just a representation of reality and you can represent it in infinitely many ways.
Sure. But only one model makes sense without distortion. If you map known distances, land mass shapes and sizes and flight routes on to a sphere they make sense.
Try and do it on a flat plane and there is distortion. You seem to be getting round that by distorting the coordinate system itself.
I'd suggest an earth where a meter in Greenland is the same as a meter in Australia is far more likely.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 02, 2022, 12:28:02 PM
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Philosophical implications:
The trick above generalizes to all coord transforms. You can practically turn any shape into any other shape.
And physics, well they use coord transforms all the time.
So all of physics basically works on any shape universe

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

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

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

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

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

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

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

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


Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 12:40:22 PM
In your above image at the left, the Sun's rays project light in only a "downward" direction from a single point.

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

Really sorry, I did indeed miss your question.
Yes you are right, the sun emits rays in all directions. I mainly wanted to show day/night on earth, and only drew rays to dusk/dawn areas. (to not clutter the picture)
Also bending of light is only strong around earth, so you wouldn't see that much bending around the sun.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 12:52:24 PM
Yes we do. It has been measured and observed and there's only one shape which makes sense with those observations and measurements.
Only if you're presupposing an orthonormal basis. But orthonormal basis is an assumption and not a law of nature.
All you've done is show that you can map the reality on to different shapes.
Yes! And therefore we can't know the true shape.

Sure. But only one model makes sense without distortion. If you map known distances, land mass shapes and sizes and flight routes on to a sphere they make sense.
Try and do it on a flat plane and there is distortion. You seem to be getting round that by distorting the coordinate system itself.
I'd suggest an earth where a meter in Greenland is the same as a meter in Australia is far more likely.
You're again presupposing an orthonormal coordinate system. Mathematically the measurements make exactly as much sense in any of the other coordinate systems. You can't take an orthonormal ruler to a non-orthonormal base. If i take my curvy earth ruler to your sphere it also wouldn't make sense at all.
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on February 02, 2022, 01:59:51 PM
Mathematically the measurements make exactly as much sense in any of the other coordinate systems.
But in physical reality, not so much.
What you have got is an interesting thought experiment or philosophical thought, nothing more.
I'm assuming that a meter in Greenland is the same as a meter in the UK is the same as a meter in Australia, and that a 90 degree angle is the same everywhere too.
I believe that's a reasonable assumption - that the scale of reality doesn't randomly change depending on where you are.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 02:25:42 PM
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A curvature such as Gaussian curvature which is detectable to the "inhabitants" of a surface and not just outside observers. An extrinsic curvature, on the other hand, is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides.

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

Could you perhaps help me understand at what point the math breaks?
- The first step is that i express everything in celestial coordinates(lat, long, distance). But I don't think this breaks curvature because earth is still a sphere and all distances are the same.
- The second step is that i draw latitude on a straight axis instead of an angular one. So numerically everything stays the same. How can the math then notice something changed? It can't see the drawing and the numbers didn't change?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 04:15:03 PM
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A curvature such as Gaussian curvature which is detectable to the "inhabitants" of a surface and not just outside observers. An extrinsic curvature, on the other hand, is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides.

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

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

I currently have two conflicting sentiments:
- Your claims about intrinsic curvature which i can't even pretend to be knowledgeable enough to assess let alone start speculating about the effect of teleporters on the math.
- The fact that there are no observable differences between the two models. You said intuitively the differences in distances and angles give the shape away, but there are no observable differences with the globe model. All distances and angles measured are the same in both models (for an inside observer)
Would you please share your thoughts?
Title: Re: Found a fully working flat earth model?
Post by: Iceman on February 02, 2022, 04:23:12 PM
Can you elaborate on this teleporting across the ‘edge’ of Antarctica in relation to curvature? What exactly do you mean?
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 02, 2022, 04:47:23 PM
Imagine i have 2D cartesian coordinates (0,0) and (4, 4).  Then the distance formula would be sqrt((x1-x2)² + (y1-y2)²
In polar coordinates these coordinates would be (0°, 0) en (45°, 4).  If i plug these numbers into the distance formula, i get total gibberish. When you do a coord transform, you must update all formulas. That is what i mean with a new distance metric. It's the old one with compensation for the coord transform.

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

As folks have been trying to tell you, distance is distance.  Changing the coordinate system doesn't change that.  It just changes how you are describing it as well as how you'd calculate it but it doesn't change what it actually is.
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 02, 2022, 05:07:21 PM
Consider this thought experiment: that shows the earth can have any possible shape ...  We will simply never know the shape of the planet.

Consider this thought experiment;

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

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

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

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

Basically; if the Earth was NOT a globe, dontcha think someone out of these hundreds of folk would have noticed ....?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 05:40:27 PM
What you have got is an interesting thought experiment or philosophical thought, nothing more.
agreed. The model will not be of much use in practice. Just the philosophical conclusion that earth can have any shape and we'll never know was rather surprising to me when i stumbled onto it.
I'm assuming that a meter in Greenland is the same as a meter in the UK is the same as a meter in Australia, and that a 90 degree angle is the same everywhere too.
I believe that's a reasonable assumption - that the scale of reality doesn't randomly change depending on where you are.
I want to clarify that for an observer inside the flat model, all these measurements will match up.
But for you as an outside observer that's indeed a very handy property  :)
One small caveat: when i'm driving to the supermarket, the flat model does match my intuition more.
My interpretation of  occams razor is: If you have 2 models, take the one that makes your job easiest, and you're quite correct, that will almost always be the globe model.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 06:03:12 PM
Can you elaborate on this teleporting across the ‘edge’ of Antarctica in relation to curvature? What exactly do you mean?
As you know we're mapping the globe model onto a flat earth. On the globe a hiker can happily walk across the southpole.
Plotting this trajectory on the AE-map the hiker will appear to suddenly teleport to other side of the disc.
Transforming with any other point than the northpole on top, will make antarctica continuous again, so this is just a mathematical artifact.
But for the effect of curvature, it does mean that the greatcircle northpole-southpole-northpole  is really a closed loop and not and open line.
I don't know enough about these advanced maths but
- distances and angles in the model all match the globe model for an inside observer. So i'm not sure where difference in curvature would come from
- Doing the bug-walking-a-circle and comparing with 2*π*R also seems to confirm it curves like a globe,
- by construction i can't readily identify when the curve-breaking would be happening.
- when we shift the axis slightly away from the southpole, it will suddenly become a close curve again. So i think this is asymptotic behaviour of a loop and i think the intrinsic curvature is still 1/R² in the limit.
- in general, math doesn't like teleporting :)

Again i'm in way too deep over my head, so i can't draw any conclusions. Just trying to match it with my intuition.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 06:10:43 PM
As folks have been trying to tell you, distance is distance.  Changing the coordinate system doesn't change that.  It just changes how you are describing it as well as how you'd calculate it but it doesn't change what it actually is.
What do you mean with distance?
I suppose we're comparing distance measured between 2 points in reality with:
- distance measured on the outside of the model with an orthogonal ruler
- distance measured on the outside of the model with a flat-earth ruler
- distance measured by someone living inside the model
- mathematically calculated distance.
Only the first method fails. All 3 others give the correct result.
I've already conceded that it's indeed a handy property for a model to be measurable with an orthogonal ruler.
However the entire purpose of the model is philosophical in nature and for that we only want to show that "it works", which it does.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 02, 2022, 06:17:23 PM
Does your model account for the anecdotal, photographic, geological and other proofs of all these missions?

Basically; if the Earth was NOT a globe, dontcha think someone out of these hundreds of folk would have noticed ....?
The model i present is the globe model transformed.
It turns out physics works regardless of shape (it's math, it only cares about the numbers) and it can be proven there is no test that can differentiate between the 2 models.
So if the globe model can account for all the evidence, so can mine.
Please have a look at https://forum.tfes.org/index.php?topic=19093.msg257726#msg257726 and https://troolon.com for more details about this trickery.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 02, 2022, 09:09:11 PM
Physics is looking at the world around us and developing a consistent explanation of what we see. Math is starting with some assumption and developing a logically consistent system. Physics is, math describes. The same physical object can be described in either cartessian or polar coordinates, but a sphere is still a sphere and a plane is still a plane. The coordinates are not bounded, 3 space to infinity is within both. The reeason locations on the globe are given as latitude and longitude, i.e. polar coordinates, is that the earth is round. If it were flat, we would use x/y coordinates, much easier.

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

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

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

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

Then there is Sigma Octatis, the southern (pretty close) pole star. At the summer (northern hemisphere) equinox, at midnight in South Africa it is just after sunset in western Australia and just before sunrise in South America. You can see Sigma Octatis directly south from all these places at that time. On FE disk map, Sigma Octatis is in directly opposite directions from South America and Australia.
 
So please show the dome appearance in your model. I will be interested to see: Where is Sigma Octatis and how do the light rays from it travel? How does the dome appear light blue for some, and for those who see it as dark, different stars. How do those stars appear to travel across the dome in different directions at the same time? Please show with your model.

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

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

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

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

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Then there is Sigma Octatis, the southern (pretty close) pole star. At the summer (northern hemisphere) equinox, at midnight in South Africa it is just after sunset in western Australia and just before sunrise in South America. You can see Sigma Octatis directly south from all these places at that time. On FE disk map, Sigma Octatis is in directly opposite directions from South America and Australia.
 
So please show the dome appearance in your model. I will be interested to see: Where is Sigma Octatis and how do the light rays from it travel? How does the dome appear light blue for some, and for those who see it as dark, different stars. How do those stars appear to travel across the dome in different directions at the same time? Please show with your model.
Same answer as above. If it works on a globe, it works on the flat earth as it is the globe model transformed. I will not be rendering out all examples you've requested. We believe this model is only of philosophical use and thus quite pointless to do so. But i've explained all formulas, feel free to investigate further. However i have rendered the south star before. Couple of notes on the picture: All points below the southpole are transformed into a circle (thank you AE map). So the yellow circle you see is the star. The colored lines are the sightlines from the 3 observers. It's easy, draw on a globe, add lightrays from source to destination, transform. The other pictures show from where the southstar is visible and the shared area of the sky the 3 observers can see (it's the area inside the yellow circle and outside the colored ones). I hope it makes sense
(http://troolon.com/wp-content/uploads/2022/02/southstar3.png)
(http://troolon.com/wp-content/uploads/2022/02/southstar1.png)
(http://troolon.com/wp-content/uploads/2022/02/southstar2.png)

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

Also i'd like to reiterate the goal here:
We've created a thought experiment that shows that:
- the shape of the universe can be changed, without affecting the physics
- it is impossible to differentiate between the different shaped models. There exists no test, observation or measurement to find a difference.
- The universe can have many different shapes and we'll never be able to prove which one. Flat and globe are both possible though. As is a simulation, or no shape or ...
And as a side-effect we've also created a fully working flat-earth model. :)
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 03, 2022, 03:41:35 AM
The haversine gives the correct distance over the surface of a sphere between two points. When you transform this onto a 2d circle as in the FAQ map, you can have each point on the same lat/long as on a sphere. But those points are no longer on a sphere where the lat/long are both polar coordinates. Latitude is now along a straight line, you can't do trig with a straight line. Lat/long no longer has the meaning that makes haversine valid. All you are ever doing is getting the same answer as you would on a sphere. But you are no longer on a sphere, haversine means nothing when it is not segments of a curved surface. Of course you get the same answer, all you did was plug the same numbers into the same formula, but the formula does not apply to the product of the transformation. The distance must be calculated as the distance between the endpoints of the arc.

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

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

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

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

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

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

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

Sorry if I repeated myself, it is the natural human tendency when people on this site can't or won't understand.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 03, 2022, 03:50:50 AM
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Quick question, what does intrinsic curvature do with teleporters? When you walk of the edge in antarctica, you appear again at the other end of the map. Would such a teleport contribute to curvature?

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

(https://i.imgur.com/yyrDvCA.jpg)

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

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

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

https://www.youtube.com/watch?v=-Il2FrmJtcQ

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

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

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

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

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

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

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

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

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

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

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

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

(https://i.imgur.com/UCOgMAC.gif)
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 03, 2022, 03:58:08 AM
Quote
Can you elaborate on this teleporting across the ‘edge’ of Antarctica in relation to curvature? What exactly do you mean?

If that is directed at me, I think you misunderstand what "parallel transport" means.  See my response to Troolon above.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 03, 2022, 10:34:16 PM
Quote
it is impossible to differentiate between the different shaped models. There exists no test, observation or measurement to find a difference.

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

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

Notice the gyroscope can't make the full loop without changing directions.
Thank you for the crash course :) It took me a while but i think i figured it out. The intrinsic curvature of my space is 1/R² and the normal test also checks out with a globe. Mathematically intrinsic curvature depends on shape and distance. I think by changing the distance metric the shape is compensated.
The easiest way to explain is probably through construction:
- start with a sphere expressed in cartesian coords x²+y²+z² = R²
- convert to spherical coords: (lat, long, R).  Note that this is a coord transform that doesn't change curvature. The sphere is still a sphere. The reason for this is because we change 3 things at the same time: the axis, all numerical coord values and the distance metric.
- Has anyone ever noticed that angles are sometimes displayed as an angle around the origin (eg like polar coords), and sometimes they're plotted on a straight axis (eg like a sine graph). This is just a representation, it does not change the math.
Now consider our celestial coordinates from before, and draw latitude on a straight axis. This will create an AE projection. However this is just a representation of the same sphere in spherical coords. The coordinates did not change. So the gauss curvature will also not change. The intrinsic curvature of this "flat" space is still 1/R².

So in summary,
- It's possible to make physics work on a flat earth representation of the universe.
- There is no test that can distinguish the flat earth representation from the globe representation
- The true shape of the planet can be flat, or it can be a globe (or a square, or ....) but we will never be able to prove it.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 03, 2022, 11:35:19 PM
Repeat with me, haversine only works on a 3d sphere, not a 2d disk. The correct formula produces the exact thing you see in the faq map and your model. No distance funny business needed.

Sorry if I repeated myself, it is the natural human tendency when people on this site can't or won't understand.
I'm sorry if i cause frustration, that's definitely not the intention, nor do i wish to be contrary without reason. I've really been thinking about the curvature argument.

I believe you may be presupposing an orthonormal axis in all your arguments. In an orthonormal axis you are right. But we're working with spherical coordinates even though they're not drawn the conventional way.
This entire discussion really boils down to: what axis must be used to represent the universe, and orthonormal is just one of infinitely many.
There are several practical reasons to prefer orthonormal, but a lot of people forget it's just one of infinitely many possibilities.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 04, 2022, 02:41:32 AM
I am not assuming an orthonormal axis, and am sorry I brought up Gauss. Here is the explanation with high school trigonometry.

The AE map is a plane, not a sphere. If all you did was change coordinate system, you did not make a FE map. When you turn a sphere into a flat surface (assuming you don't want to map points on the sphere to the edge or bottom of FE, thus a cylinder), you are doing a projection of 3d onto 2d. Longitude lines stay the same (preserve distance), and latitude turns into radius, the angle turns into distance from north pole. If you take the north pole as 0 and south pole as 180, the formula is: Length of an Arc = θ × (π/180) × r. So (lat, longitude, radius) turns into (latitude, radius = arc length of distance from north pole). Radius and latitude turn into just radius through the arc length formula.

On RE, the longitude lines are widest at the equator and get closer together as you move towards the poles. On FE, they get farther apart in the southern hemisphere.

Fun fact: On the AE map, distance along the longitude lines is the same as RE. Distance along latitude lines gets bigger than observed south of the equator. So if you want to "fix" this by making distance flexible by latitude, you have to explain why the east/west distance gets bigger, while the north/south distance stays the same. North of the equator, the east/west distance gets smaller the farther north you go. So go to Australia with a ruler, turn it north/south, then easy/west. See if it changes length. Or car odometer? Surveyor transit? You are going to need a lot more explanation than stretchy bendy rulers.

Epistemology: Theoretically, you can never prove the earth is round. But you can 1. establish a working truth (the one that works for navigation is RE), and 2, prove that something is false.

Occam's razor: the true explanation is the simplest. For FE to be true, light has to bend due to "unknown forces with unknown equations", gravity is all messed up, a million things. For RE, all you need is the known behavior of light and physics.

So until you can place sigma octatus on your model such that it appears directly south everywhere in the southern hemisphere, show how the dome appears daylight for some, night for others at the same time, with different stars in northern/southern hemisphere, and completely different star trails from every point, I am concluding AE is falsified. Meanwhile, RE explains all that and more, consistent with known confirmed physics and observations.

Your models prove only that you can distort reality with mathematical transforms and make graphics showing anything, but none of them "work". Show me one where you see sigma octatus only in the southern hemisphere and polaris only in the northern hemisphere, both at an angle of inclination equal to your latitude, which is consistent with light traveling straight and the known laws of physics. RE works.

The only thing you can know in absolute terms is that you exist to have the thought. But for day to day life, RE works, FE doesn't.
Title: Re: Found a fully working flat earth model?
Post by: Tom Bishop on February 04, 2022, 02:50:25 AM
Occam's razor: the true explanation is the simplest. For FE to be true, light has to bend due to "unknown forces with unknown equations", gravity is all messed up, a million things. For RE, all you need is the known behavior of light and physics.

Ha, no. You just need to look out your window at the Moon and celestial bodies to know that the straight line nature of light on a celestial scale is incorrect.

https://wiki.tfes.org/Moon_Tilt_Illusion

(https://wiki.tfes.org/images/c/c4/Moon_Tilt_Diagram.png)

This curving of light also applies to the Milky Way, Aurora Borealis, Ecliptic, the Tails of Comets and other phenomena, as documented here - https://wiki.tfes.org/Celestial_Sphere

It seems to be a massive unexplainable anomaly in the sky which affects multiple celestial elements that no one can quite put their finger on.

The general argument against this is that it's some kind of vague illusion with some kind of mechanism to cause this to occur and that we should dismiss discussion of it, despite that it is what would happen if light were not traveling in straight lines on large scales.
Title: Re: Found a fully working flat earth model?
Post by: Gonzo on February 04, 2022, 07:44:17 AM
Sigh.

It’s called the ‘moon tilt illusion’ because it’s just that, an optical illusion. The illuminated side of the moon appears to point away from the sun.

As you know, if you use a straight object, or a taught piece of string, and hold it up joining the moon to the sun, you will very quickly and simply discover it does indeed point at the sun.

Perhaps you were doing it wrong?
Title: Re: Found a fully working flat earth model?
Post by: AllAroundTheWorld on February 04, 2022, 09:37:27 AM
Sigh.

It’s called the ‘moon tilt illusion’ because it’s just that, an optical illusion. The illuminated side of the moon appears to point away from the sun.

As you know, if you use a straight object, or a taught piece of string, and hold it up joining the moon to the sun, you will very quickly and simply discover it does indeed point at the sun.

Perhaps you were doing it wrong?
I can't tell if Tom is trolling or if he really doesn't understand that an optical illusion, by definition, means that things appear differently to the reality.
He posted a diagram which shows a curving line which is the apparent path the light must take from the sun to the moon.
And the point is the curve is an arch, like a rainbow which goes up and down as you look.
His argument about the string experiment is that the light could be curving away from you in an arch and that would line up with the string.
I mean, it could, but ​that isn't how the curve appears. The more relevant question is can you stretch a taught string in the direction opposite to the way the arrow is pointing and it intersect the sun.
On that diagram no you can't. In real life yes you can. I've done it. I actually got a decent picture of the illusion recently

(https://i.ibb.co/rsXrcV2/MoonTilt.jpg)

Full size image here: https://i.ibb.co/JQSJy9j/MoonTilt.jpg

I did the experiment and my string (actually a shoe lace, I hadn't gone out with the intention of doing experiments!) perpendicular to the terminator intersected the sun.
I've marked on the photo the apparent direction of the illumination. It doesn't look as though a line pointing in that direction could intersect the sun...but it does.
Again, that's what an optical illusion is.
Interestingly, I noticed that if I moved my head side to side quickly it appeared that the string itself was curving, which it obviously wasn't as it was taut.
This proves that the apparent curve is simply an optical illusion

I note that in Tom's diagram the light bends downwards, not upwards as claimed by EA.
So this phenomenon which Tom seems to think is a point for FE is maybe an experiment we can do to distinguish FE from RE.
If the earth is a globe with a moon orbiting it, the sun is distant and illuminates both and light travels in straight lines then you would expect string to line up.
And it does.

Tom is using an optical illusion to try and score a point for FE. It's the logical equivalent of claiming that parallel lines don't maintain a consistent distance and then using this optical illusion as evidence.

(https://i.ibb.co/4S6m5Fh/Hering-Illusion.jpg)

The equivalent to the string in this case is a simple removal of the background which causes the illusion

(https://i.ibb.co/PFZRzSs/Hering-Illusion-Broken.jpg)

(Or you could use a string to satisfy yourself that the lines are indeed straight.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 04, 2022, 01:57:07 PM
Quote
The intrinsic curvature of my space is 1/R²
How did you calculate that?
Quote
If one of the principal curvatures is zero: κ1κ2 = 0, the Gaussian curvature is zero and the surface is said to have a parabolic point.
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The Gaussian curvature is the product of the two principal curvatures Κ = κ1κ2
.
https://en.wikipedia.org/wiki/Gaussian_curvature#Gauss%E2%80%93Bonnet_theorem
Where are your principle curvatures?  It doesn’t look to me like there are any curvatures.

Quote
Imagine i have 2D cartesian coordinates (0,0) and (4, 4).  Then the distance formula would be sqrt((x1-x2)² + (y1-y2)²
In polar coordinates these coordinates would be (0°, 0) en (45°, 4).  If i plug these numbers into the distance formula, i get total gibberish. When you do a coord transform, you must update all formulas. That is what i mean with a new distance metric. It's the old one with compensation for the coord transform.
What “formula” did you use?  Coordinate systems use metric tensors to fix distances and different coordinate systems have a defined metric tensor that should be used. 

-
Quote
There is no test that can distinguish the flat earth representation from the globe representation

You are contradicting yourself.  You claim your model has intrinsic curvature, but, by definition, intrinsic curvature can be tested for and observed from “the inside”
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 04, 2022, 02:59:45 PM
You just need to look out your window at the Moon and celestial bodies to know that the straight line nature of light on a celestial scale is incorrect.

... but it's perfectly straight at this scale?

(https://wiki.tfes.org/images/1/1a/Experiment-2a.jpg)
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 04, 2022, 05:05:03 PM
Haven't thoughtfully tried it yet, but I'd propose the easiest way to debunk Tom's moon tilt defense is to observe the moon in the dark either pre dawn or post sunset in order to know where the sun should be.  In the dark, I would think the background causing the illusion would be removed.  Not sure that's true or not.  Just a thought.

I drive to work pre dawn every day.  I did notice last week that the moon tilt was definitely directed at the sun. However, I need to get a clear morning when the moon is close to setting to really see what the moon tilt looks like across the entire sky.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 04, 2022, 05:20:44 PM
Tom Bishop,

I have some questions about EA, will post soon, perhaps next since you brought it up. Star trails and the apparent motion of astronomical objects as observed by someone on RE which is spinning and tilted is not at all the same thing as the question of whether light travels in straight lines in a vacuum. You do not see the path of the light in a vacuum. The tail of a comet and the path of a comet could appear curved, but that is not the same thing as the light traveling in a curved path.

So where on the celestial dome is sigma octatus?

How does it appear directly south of all points in the southern hemisphere? When it is sunset in Denver, how do people in Salt Lake City see the entire dome as light blue, while people in St Louis see it as dark with stars? How does someone in St Louis see right through the daylight to the dark sky with stars past the Salt Lake CIty daylight? At that time, how can someone in Salt Lake City and someone in St Louis look at the same point in the sky over Denver and one sees daylight while the other sees dark with stars? How do people in the southern hemisphere see stars over the entire dome while someone in northern hemisphere sees completely different stars over the same dome, with startrails different from every point, while other people see light blue daytime sky, same time, same dome?

Seems to me, the bending must go every which way depending on where you are and what you are looking at. Can you diagram the light paths to account for St Louis seeing dark and stars over Denver at the same time that Salt Lake City sees daytime light blue at the same spot?

RET explains this, diagrams available in science textbooks. Is there a corresponding diagram and explanation for FE?

Where is sigma octatus?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 08:33:17 PM
The AE map is a plane, not a sphere. If all you did was change coordinate system, you did not make a FE map. When you turn a sphere into a flat surface (assuming you don't want to map points on the sphere to the edge or bottom of FE, thus a cylinder), you are doing a projection of 3d onto 2d. Longitude lines stay the same (preserve distance), and latitude turns into radius, the angle turns into distance from north pole.
The surface of a sphere (shell) is a 2D object. Any point on it can be represented by 2 coords(lat, long). A disc is also 2D. No loss of dimensions or information. A sphere is 3D. In my model, the sphere is transformed into a cylinder, also 3D.
In formulas: on a sphere: coords are expressed as (lat, long, radius), On the cylinder they're expressed as (lat, long, height) (lat is rendered on a straight axis, height from bottom which equals radius).
The only difference between between the sphere and the cylinder is wether we draw latitude on a straight, or a radial axis. Math uses both representations interchangeably (eg: polar coords -> radial, sine graph -> straight).

Personally I prefer to express Occam's razor as: "When you have multiple models, choose the one that's easiest for the problem at hand". Physics continuously switches between flat, spherical, relativistic and QM models depending on the problem.

Globe is just the dominant way to represent physics, bit it's just a representation. Cylindrical is another one.

BTW, here's an observer traveling the globe and showing what part of the sky is visible. It explains from where you can see polaris and the south star. (i already uploaded pics on how different people can see the south star from different places)
(http://troolon.com/wp-content/uploads/2022/01/horizon_cone.gif)
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 09:00:41 PM
Quote
The intrinsic curvature of my space is 1/R²
How did you calculate that?
1. By construction. If a sphere in spherical coords has curvature 1/R², then so has mine
2. by the bug walking a circle test and comparing circumference of the circle with 2πR
3. by the fact any distance between any 2 points on the flat earth with my distance metric, is the same as on the globe. No distance has changed, then how can the intrinsic curvature?
4. By flying a plane with the normals test
Not all coord transforms change curvature (eg: cartesian sphere: x²+y²+z²=R² versus celestial sphere: lat=[0-π], long=[0-2π], distance=R)
All i'm doing is representing latitude on a straight axis instead of a radial one. This doesn't change the intrinsic mathematical shape.
It's like claiming i broke physics because i drew my graph in red pen instead of a black one.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 09:26:34 PM
You just need to look out your window at the Moon and celestial bodies to know that the straight line nature of light on a celestial scale is incorrect.

... but it's perfectly straight at this scale?

Straight like this? :)
(http://troolon.com/wp-content/uploads/2022/01/horizon-curve.png)
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 09:31:27 PM
RET explains this, diagrams available in science textbooks. Is there a corresponding diagram and explanation for FE?

Where is sigma octatus?
Yes, check my model. Anything physics can explain, my model can do because it's the same model.
Physics works regardless of shape.
FE proof = express everything in spherical coords; proof in RE; express result in celestial coord again.
(or you can derive all formulas by hand, but that will take a while longer)
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 09:34:41 PM
Where is sigma octatus?
In my model: express sigma octatis as (lat, long, distance) from the center of the earth
Draw lat/long as on a AE map and place that in a cylinder at height `distance`
ie long on a radial axis, lat on a straight axis, distance in the up direction.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 04, 2022, 10:11:37 PM
Troolon says: In my model: express sigma octatis as (lat, long, distance) from the center of the earth

The RE position of sigma octatus: Directly over latitude 90 degrees south at a distance of 280 million light years Longitude makes no sense, since all longitude lines come together at the south pole. It is 280 million light years in the direction of the axis of rotation.

                * Polaris



              _|_
             /     \
            (-----)       Earth (sorry for crude graphics, not going to take the time to find and learn graphics etc
             \___/       Axis is vertical lines, equator is dashed line
                |




               * Sigma Octatus

In this model, Sigma Octatus is in one place, visible to all south of the equator, always appearing directly south to all.

On the AE map, Sigma Octatus appears to each person in the southern hemisphere as being directly away from the center projected out radially along the longitude line you are standing on

AE map:

                     * Sigma Octatus viewed from Australia

                   ____
                  /       \
                 (         )              * Sigma Octatus viewed from South Africa
                  \____/


                     * Sigma Octatus viewed from South America

Please explain where Sigma Octatus is.
Title: Re: Found a fully working flat earth model?
Post by: Gonzo on February 04, 2022, 10:16:01 PM
Troolon,

In your diagramme above, what would I experience if I was stood on the top of the tower block, looking at the person on the right? Would they appear to be a lot lower than they actually were in reality? (Let’s remove the boat for the moment)
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 04, 2022, 10:46:16 PM
Further thought re the location of Sigma Octatus in the FE dome model. Seems like it would be on the dome, so ...

For someone at the equator, Sigma Octatus appears on the horizon directly away from the north pole, which would put it on the ground everywhere along the edge. For someone at the south pole, it appears to be at the center of the dome, and directly above you.

So on the AE map, Sigma Octatus is everywhere on the dome from the center top to everywhere on the edge.

Looking forward to seeing how your model incorporates that. Where is Sigma Octatus? RE has consistent plausible location. FE does not.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 04, 2022, 10:59:34 PM
Also, in your model, you show the light bending such that it crashes into the ground where sunset/sunrise occurs. Do you have an explanation why another light ray coming off the sun at slightly higher angle would not hit the ground beyond the sunset/sunrise line? What limits the direction the rays come off the sun? If nothing does, why is the whole earth not illuminated 24 hrs/day?

You need curved light AND a "lampshade" or directional beam sun.


Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 11:25:42 PM
Please explain where Sigma Octatus is.
On the AE map, where is the southpole? It gets deformed into the circle around the edge of the disc.
In my map, the southstar also gets deformed into a circle. Once you have a latitude that's even slightly off of -90°, it will become a point again.
So in the image below, the southstar is the yellow circle, and you can see all 3 observers see the circle due south.
You asked me to drarw the most abhorrent case on the AE cylinder, you shouldn't be surprised it looks weird :)
(btw on the globe, the southpole has no unique coordinates: lat=-90, long=[0-2π] which explains this result)

Once again this FE model can explain anything the globe model can because... it is the globe model.
(http://troolon.com/wp-content/uploads/2022/02/southstar3.png)
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 11:35:23 PM
In your diagramme above, what would I experience if I was stood on the top of the tower block, looking at the person on the right? Would they appear to be a lot lower than they actually were in reality? (Let’s remove the boat for the moment)
The light curves exactly the opposite way the earth would have curved on a globe.
On a globe, 2 buildings are not parallel, they slightly veer out due to the curvature, and so on a globe you would also look slightly down to see the other person.

     \ __ /          <- two very tall buildings not being parallel due to curvature.       
     /     \           <- wannabe globe ;)
      \__/
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 04, 2022, 11:45:15 PM
You need curved light AND a "lampshade" or directional beam sun.

In reality you can't see "below" the horizon. This is why ships disappear keel-first.
When you create the horizon-equation (horizon here meaning the line you can't see below), you will see it curves up towards infinity in all directions, creating a cone around the observer.
As you can't see below the horizon, you also can't see outside the cone.
From the viewpoint of the sun, light travels in all angles towards the earth. However the curvature of the light on the non-lit hemisphere, is so extreme the light curves back into space without hitting the earth. There's no need for lampshades when curving makes the light miss the earth.

BTW, as an answer to all your questions:
- draw your question as the globe prescribes (ie draw lightrays a sun and a earth).
- express all elements in polar coords
- draw as a flat earth. (ie put latitude on a straight instead of a radial axis)
It all works because this model IS the globe model.  Anything the globe can explain, so can this model...
Physics cares about numbers, not shapes.
We'll never know the shape of the earth (if it even has a shape)
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 05, 2022, 12:09:33 AM
On a globe, 2 buildings are not parallel, they slightly veer out due to the curvature, and so on a globe you would also look slightly down to see the other person.

     \ __ /          <- two very tall buildings not being parallel due to curvature.       
     /     \           <- wannabe globe ;)
      \__/

Yup, stand at the top of one building, align a spirit level or similar to show horizontal, and the line drawn along that level does not meet the building of similar height some distance away, as you would expect it to do if the Earth were flat.

Or, stand at a set height (210m in one example), look at an object of 210m some km distant, and find that the sightline between the two does not meet the hills of 400m+ behind at 210m, where it should if the Earth were flat, but passes clear over them.

Practical stuff that shows, regardless of your model, that the Earth is Not Flat.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 12:18:52 AM
Yup, stand at the top of one building, align a spirit level or similar to show horizontal, and the line drawn along that level does not meet the building of similar height some distance away, as you would expect it to do if the Earth were flat.
If light curves upward, that would be rather logical.
My model is the globe model coord transformed. (if even that). So if the globe can explain it, so can mine.
I've taken the globe model, removed the curve and contorted anything to make the math match.

Or, stand at a set height (210m in one example), look at an object of 210m some km distant, and find that the sightline between the two does not meet the hills of 400m+ behind at 210m, where it should if the Earth were flat, but passes clear over them.
light curving explains that as well. The light curves in such a way it compensates the missing curvature of the earth

Practical stuff that shows, regardless of your model, that the Earth is Not Flat.
This model is nothing more than mathematical trickery.
There is no test to differentiate the two models. And if we have 2 different shaped models, that can explain everything, we can't ever be sure of the shape of the earth.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 05, 2022, 01:10:41 AM
Troolon,

The curvature of a plane (FE) is zero. Whatever the curvature of a sphere is, it is not zero, that's all you need to know for Gauss. Draw a map on a piece of paper, then wrap it around a globe. You will have to tear or wrinkle it, that represents distance errors. If you draw the map on a paper wrapped globe without tears or wrinkles, then spread it out on a flat surface, again tears and wrinkles, thus distance errors because when you change the curvature, distances change. WHen you can peel an orange and then flatten the peel without tearing it, you will be able to make FE map. Never gonna happen.

Distance on RE vs FE/AE map. Consider Sydney (33°52′S 151°12′E) to Santiago (33°27′S 70°40′W), famous airline route much discussed in FE community.

RE, calculated using haversines or equivalent trig:

Google maps says distance is 7078 miles.
Lat/long distance calculator says 7153 (I didn't put in the exact lat/long).
Airline schedule says 7043 miles.

FE/AE:

Draw a line from north pole to Sydney, and a line from north pole to Santiago. then draw a line from Santiago to Sydney. Now you have a triangle with angle at the north pole of lat2 - lat1, in this case 139 degrees. Distance from north pole (one side of triangle) is angle between north pole and 33 degrees south, which is 123/180 * 12,500 or 8541 miles. Now we have a triangle with two sides = 8541 and the angle between these two sides is 139 degrees. the base is the distance between Santiago and Sydney. Using triangle calculator (lazy, remember?), I get 16,000 miles.

If the distance is about 7000 miles, the earth is round. If the distance is 16,000 miles, the earth is flat. There is an airline flight, they fly at 500 mph. On FE, this flight would be 32 hours long and far longer than the range of any airliner.

On RE, you need spherical coordinates and haversines (or equiv trig). On FE/AE, it is easy to make a 2d triangle and all you need is elementary 2d geometry. The same equations do not work on both, on FE there is no longitude angle because it is a straight line. No r is available on FE latitude, so not hversines. Say it with me, there is no r on latitude on FE, can't use haversines.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 05, 2022, 01:20:34 AM
I could write an equation to do a transform of globe to a single point. You keep saying it is all the same, but it isn't. You can figure out distances on FE with triangles. You can't do that on RE, you need trig. Your transform goes from a sphere to a plane, the very definition of FE. The distance equations are different on a sphere and a plane.

I guess you want to have your idea more than you want to do math right.

Australia is bigger east/west on AE map, but the same as RE for north/south. That is not possible.

Proof by contradiction: assume the earth is flat. Observe this leads to a map with wrong distances. QED, FE is falsw.

Title: Re: Found a fully working flat earth model?
Post by: jimster on February 05, 2022, 02:16:53 AM
If someone is at 80 degrees south latitude, where is Sigma Octatus? It appears as a point 10 degrees south of directly overhead for observers anywhere on that latitude. For observers at 70 degrees south, it appears 20 degrees south of directly overhead. The points at each latitude will form a circle, so you need lots of circles, not just one. If you had observers over the entire hemisphere, Sigma Octatus would cover the entire dome. It appears at every point around the circumference and at every angle of inclination.

Pick any point on the dome and you can calculate where an observer would be to see it there. The observer would be on a line between between the point you picked and the north pole, They will be at a latitude equal to the angle below straight up. Sigma Octatus must be everywhere. You get a circle for every latitude.

How can it be a circle? Does the circle appear as a star? Even at 10 degrees south, observers on opposite sides of the globe will see it in different places. Sigma Octatus is not a circle. It is a small dot. And on FE, it is in completely different directions depending on where you are.

On RE, south pole is a point. You say it is a giant circle. That is a deformation that proves your model is not equivalent to RE. The model can't be equivalent when the south pole is a point on RE a giant circle on AE. That is huge distortion.

Not just a problem at 90 degrees south, a problem everywhere in the southern hemisphere. Pick any point on the dome and I can show you a place where it will appear to be there. On FE, Sigma Octatus must be everywhere.

I keep repeating myself as though the problem is you didn't hear me, when the real problem is called motivated reasoning - you will not accept that FE is falsified.

If you take your theories to mathemeticians or astronomers, they will say the same as me. Perhaps you are right and all mathematicians and astronomers are wrong. Maybe you are a genius like Newton.
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 05, 2022, 03:43:28 AM
I think the problem is twofold, jimster. First, you can't spell Sigma Octantis at all and apparently refuse to even check. Second, and probably more importantly,  you haven't bothered actually reading anything troolon has written about the model, because had you done that, you'd at least recognize the fact that objects map identically in the standard RE framework as they do in troolon's FE model. You've completely ignored the distance metric that is widely different than the ruler you are used to as well as the coordinate transformation. The knee-jerk reaction people seem to have to anyone that even gives a hint that they might be non combative toward FE is astounding to behold.

Troolon, I admire your patience. This is a neat construction you've presented here, and clearly done a lot of work to flesh out and demonstrate.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 05, 2022, 05:21:50 AM
Quote
It's like claiming i broke physics because i drew my graph in red pen instead of a black one.
Its a lot more complicated that that
Quote
I believe everything works because of the distance metric changing. My guess is that the changes to the distance metric nullify the changes in shape and i believe i can make a compelling case for this.

The question is did you transform the metric correctly? It is the metric, not the coordinate system, that defines curvature. It is the metric that measures how much the object deviates from flat by giving you distances and angles. The metric tensor exists independent of a coordinate system and is an inherent property of the object. Changing the coordinate system doesn’t change the properties of the object. You are just changing the environment it is in. You are correct about that.
But  when you do a coordinate transformation, you have to transform the metric tensor as well.   Actually, you just transform the components, but the tensor as a whole does not change.  But you can’t just do that willynilly and just make up any arbitrary components  or metric that you want.  That seems to be what you have done. There are rules about how to transform the metric.
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The most general transformation will transform coordinates, the metric tensor, the torsion tensor and also all other fields defined on the manifold. In case when metric, torsion and other tensor fields transform accordingly to coordinates the way tensor fields should transform, we have an coordinate transformation and manifold and all ''geometry'' (and physics) is preserved. In case metric and torsion transforms accordingly but other physical quantities does not, you are probably changing physics but preserving manifold. In case torsion or metric transformed differently than they should based on tensor transformation laws, you have defined new manifold.

Source https://www.physicsforums.com/threads/metric-tensor-transformations.584724/ .
Translation: Transforming the metric tensor correctly is critical to preserving the physics and geometry. It isn’t just about the physics working in any coordinate system.  Unless you transformed the metric tensor based on the tensor transformation laws, you have changed the geometry of the underlying structure.

When you correctly transform the metric when transforming from a spherical to a flat coordinate system, you get distortion, as explained here.
(https://i.imgur.com/oIsZui8.png)

https://phys.libretexts.org/Bookshelves/Relativity/Book%3A_Special_Relativity_(Crowell)/07%3A_Coordinates/7.03%3A_Transformation_of_the_Metric
Quote
The procedure employed above works in general. To transform the metric from coordinates (t,x,y,z)(t,x,y,z) to new coordinates (t′,x′,y′,z′)(t′,x′,y′,z′), we obtain the unprimed coordinates in terms of the primed ones, take differentials on both sides, and eliminate t,...,dt,...t,...,dt,... in favor of t′,...dt′,...t′,...dt′,... in the expression for ds2ds2. We’ll see in section 9.2, that this is an example of a more general transformation law for tensors, mathematical objects that generalize vectors and covectors in the same way that matrices generalize row and column vectors. .
You broke physics by making up your own rules for transforming the metric tensor. If you don't have any distortion in your model, you didn't transform the metric correctly and have changed the geometry.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 09:54:50 AM
The curvature of a plane (FE) is zero. Whatever the curvature of a sphere is, it is not zero, that's all you need to know for Gauss. Draw a map on a piece of paper, then wrap it around a globe. You will have to tear or wrinkle it, that represents distance errors.
See https://forum.tfes.org/index.php?topic=19093.msg258073#msg258073 .
As i've tried to explained many times, my model is a sphere in celestial coordinates.
But instead of drawing latitude on a radial axis, i draw it on a straight one. As earth radius is so unfathomably big, this is not an unreasonable assumption even...
The drawing is just a representation, it does not change the math.
You're trying to prove that a sphere loses it's curvature when it's expressed in celestial coords

If the distance is about 7000 miles, the earth is round. If the distance is 16,000 miles, the earth is flat. There is an airline flight, they fly at 500 mph. On FE, this flight would be 32 hours long and far longer than the range of any airliner....
This is a very disingenuous argument. I've said countless times before, you should use celestial coordinate geometry (ie haversine et al) to calculate distances.
What you're doing is akin to taking my model, changing all the formulas and then saying it doesn't work.
My model is the globe model, expressed in celestial coords. Please don't use sqrt(x²+y²+z²) for celestial coords

On RE, you need spherical coordinates and haversines (or equiv trig). On FE/AE, it is easy to make a 2d triangle and all you need is elementary 2d geometry. ...
My model is working in celestial coordinates. It's just not drawn the conventional way, but it are celestial coords.
Of course 2D geometry will break on celestial coords.
You're trying to convince people to use cartesian formulas on celestial coords. This reasoning would break half of physics....
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 05, 2022, 09:55:44 AM
if we have 2 different shaped models, that can explain everything, we can't ever be sure of the shape of the earth.

If you find yourself at this kind of impasse, you need to look outside your models, then. ("This model is nothing more than mathematical trickery.", you said)

Take yourself away from your desk, go out into the world, and gather some data.

Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 10:07:23 AM
I could write an equation to do a transform of globe to a single point. You keep saying it is all the same, but it isn't. You can figure out distances on FE with triangles. You can't do that on RE, you need trig. Your transform goes from a sphere to a plane, the very definition of FE. The distance equations are different on a sphere and a plane.
Yes! A coord transform can change any shape into pretty much any other shape (not a point though, that would break the physics)
But in physics you have to transform the formulas together with the coordinates.
In cartesian space, distance is calculated with √(x²+y²+z²). When you switch to celestial coords or polar coords or ...  this formula goes out the window. You have to use the appropriate distance formula for your coordinate system.
Your claim that i have to use the cartesian distance formula, on spherical coordinates will indeed break everything. Of course i can see Australia doesn't measure on the AE map in cartesian distances. The only conclusion we can draw is that cartesian distances are the wrong formula on the AE cylinder.

I guess you want to have your idea more than you want to do math right.
The math is right. I wouldn't have published it if it weren't first scrutinized by various physicists...

Proof by contradiction: assume the earth is flat. Observe this leads to a map with wrong distances. QED, FE is falsw.
To correctly state your proof:
Assume the earth is flat and we're using the cartesian distance formula.
Observe the leads to a map with wrong distances
-> conclusion: flat earth can not work with cartesian distances
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 10:12:34 AM
If someone is at 80 degrees south latitude, where is Sigma Octatus? It appears as a point 10 degrees south of directly overhead for observers anywhere on that latitude. For observers at 70 degrees south, it appears 20 degrees south of directly overhead. The points at each latitude will form a circle, so you need lots of circles, not just one. If you had observers over the entire hemisphere, Sigma Octatus would cover the entire dome. It appears at every point around the circumference and at every angle of inclination.
I can't say it any better than Clyde Frog
https://forum.tfes.org/index.php?topic=19093.msg258141#msg258141
It is the RE model you're debunking...
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 11:36:12 AM
I think the problem is twofold, jimster. First, you can't spell Sigma Octantis at all and apparently refuse to even check. Second, and probably more importantly,  you haven't bothered actually reading anything troolon has written about the model, because had you done that, you'd at least recognize the fact that objects map identically in the standard RE framework as they do in troolon's FE model. You've completely ignored the distance metric that is widely different than the ruler you are used to as well as the coordinate transformation. The knee-jerk reaction people seem to have to anyone that even gives a hint that they might be non combative toward FE is astounding to behold.

Troolon, I admire your patience. This is a neat construction you've presented here, and clearly done a lot of work to flesh out and demonstrate.
Thank you very much.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 12:34:02 PM
Quote
It's like claiming i broke physics because i drew my graph in red pen instead of a black one.
Its a lot more complicated that that
My transformation is twofold. Can you please help me understand which of these 2 steps you believe breaks curvature:
- convert everything to celestial coordinates (lat, long, distance)
  This transformation is used in physics all the time and just expressing a sphere in celestial coords doesn't break distances or curvature. (the distance formula does of course get transformed along with the coordinates, just like when physics switches to/from celestial coords)
- Please have a look at the two graphs below. One in "polar coordinates", the other one in "cartesian coords".
  These are both visualizations of the same function, In the "polar" visualization, the X axis is radial around the origin. In the "cartesian" rendering it's a straight line. Both graphs represent the same function and mathematically the functions curvature doesn't change, when plotted differently. From here stems the red/black pen analogy. FE/RE is just a representation of the same physics/mathematics.
 
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The question is did you transform the metric correctly? ....
I believe so. I would use the same distance metric physics normally uses when switching to celestial coords.
Or more formally:  distance_in_celest(p1, p2) = distance_in_cart(celest_to_cart(p1), celest_to_cart(p2))
So basically same formula after appyling the inverse transform.
I'm not proposing to do the celestial coordinate transformation any different than physics does it today.

Quote
When you correctly transform the metric when transforming from a spherical to a flat coordinate system, you get distortion, as explained here.
You're now sneakily going back to an orthonormal basis. When you take a non-orthonormal basis (like celestial coords) and start measuring it with a "straight" ruler, indeed nothing will match up. If i take my curved polar-coordinate-ruler to your globe, everything will be broken too.

Quote
You broke physics by making up your own rules for transforming the metric tensor. If you don't have any distortion in your model, you didn't transform the metric correctly and have changed the geometry.
Take the cartesian point (1,1).  distance to the origin is √x²+y² or    √2.
Express the point in polar coords: (45°, √2). The original distance formula is now invalid. The correct formula (in general) would be:
    distance_in_polar(p) = distance_in_cart(polar_to_cart(p))

So in summary i agree with all the math you've presented. However the distortion only appears once you treat the transformed coordinates as orthonormal.
In my model you do indeed lose the possibility to measure distances with a "straight" ruler.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 12:37:01 PM
if we have 2 different shaped models, that can explain everything, we can't ever be sure of the shape of the earth.
If you find yourself at this kind of impasse, you need to look outside your models, then. ("This model is nothing more than mathematical trickery.", you said)
Take yourself away from your desk, go out into the world, and gather some data.
There is no way to differentiate the models. No test. You may gather any data you like, and both models will return the same prediction each and every time.
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 05, 2022, 05:34:55 PM
There is no way to differentiate the models.

Why? Because you say so? Gotta have something better than that....
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 05, 2022, 05:45:40 PM
There is no way to differentiate the models.

Why? Because you say so? Gotta have something better than that....
Because one is simply a mapping of the other into a different coordinate system. They are the same.
Title: Re: Found a fully working flat earth model?
Post by: poweraide on February 05, 2022, 08:23:20 PM
if we have 2 different shaped models, that can explain everything, we can't ever be sure of the shape of the earth.
If you find yourself at this kind of impasse, you need to look outside your models, then. ("This model is nothing more than mathematical trickery.", you said)
Take yourself away from your desk, go out into the world, and gather some data.
There is no way to differentiate the models. No test. You may gather any data you like, and both models will return the same prediction each and every time.

Show a wave expanding from a non-central point and returning to the same point in the proposed FE model.


 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 08:46:39 PM
Show a wave expanding from a non-central point and returning to the same point in the proposed FE model.
ScienceItOut made a video showing that:
https://www.youtube.com/watch?v=s2pPCrCgvbQ
But you might be missing the point: If an expanding wave can be explained by the globe model, so it can in my model. They're equivalent.
Title: Re: Found a fully working flat earth model?
Post by: poweraide on February 05, 2022, 08:50:38 PM
Can they? Because on a flat earth, they'd have to reflect off something. There's no connection between one edge and the other, correct?

And once waves reflect, they won't return to the same spot, unless the wave starts from the very center of the transformation.
Title: Re: Found a fully working flat earth model?
Post by: stack on February 05, 2022, 09:04:23 PM
I'm not sure this is entirely relevant, but it kind of looks similar to the creation of the model. Troolon, you may (or may not) find this interesting:

Transforming from Geographic to Celestial Coordinates (http://mcellin.me.uk/artfulcomputing/images/Projects/CoordinateConversion.pdf)

The simplest astronomical observation of all is that the stars appear to move around the Earth (which, of course is the result of the Earth rotating with respect to the ‘fixed’1 stars). We therefore catalogue the position of astronomical bodies against a ‘Celestial’ coordinate system that is fixed with respect to the stars, but moves around with respect to Earth based coordinates.

The plan will involve four stages:
1. Convert to a cartesian (‘xyz’) coordinate system, with x pointing north, y pointing west, and z to the sky. (Y to the west gives us a right-handed coordinate system, which is the same as the celestial coordinate system.)
2. Rotate the axes of our coordinate system around the east-west direction so that the z axis is now parallel with the rotation axis of the Earth.
3. Rotate the axes around the new z direction (we will call this z 0) until the x axis points towards the vernal equinox, which is the point on the sky where in its annual progress around the ecliptic plane the Sun crosses the celestial equator moving northwards (usually about 20th of March).
4. Convert back from cartesian to polar coordinates to get the celestial values.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 10:20:20 PM
I'm not sure this is entirely relevant, but it kind of looks similar to the creation of the model. Troolon, you may (or may not) find this interesting:

Transforming from Geographic to Celestial Coordinates (http://mcellin.me.uk/artfulcomputing/images/Projects/CoordinateConversion.pdf)
Awesome, thanks!
I was already looking for a way to convert RA/dec to lat/long at a given time. If only to make the simulation more realistic.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 05, 2022, 10:26:27 PM
Can they? Because on a flat earth, they'd have to reflect off something. There's no connection between one edge and the other, correct?

And once waves reflect, they won't return to the same spot, unless the wave starts from the very center of the transformation.
The boundary of the disc is nothing but a mathematical artifact. Things crossing the southpole, appear at the other end of the map.

But as a general answer, my 3D model is globe physics, just rendered differently. If you take a globe, express it in in celestial coords (lat, long, distance), and then draw latitude on a straight instead of a radial axis, you'll get a flat-earth-universe. Physics works with celestial coords, and so it behaves like a globe, it just looks flat. So if a wave works on a globe, it also works on the flat renderering. The only difference is the shape of one axis.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 06, 2022, 07:51:22 AM
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flat earth can not work with cartesian distances
This is exactly what you are not getting.  There is no such thing as “cartesian distance” or “spherical distance” or “polar distance”.  When you transform the metric tensor correctly, the distances are the same no matter what coordinate system you use
Quote
The distance between two points will not change if you change your coordinates. Also the metric tensor itself, as a function, will not change. But the matrix which represents the metric tensor depends on what coordinate system is being imposed on all of the tangent spaces, so the matrix representation of the metric tensor will in fact change with changes of coordinates.
The whole point of coordinate transformations is that you don’t change the physical nature or geometry of the underlying structure.  If your transformation results in different distances between points and you have to use a "bendy ruler" to make the distances work, that is exactly what you have done.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 06, 2022, 04:51:11 PM
the distances are the same no matter what coordinate system you use
agreed
Quote
The whole point of coordinate transformations is that you don’t change the physical nature or geometry of the underlying structure.  If your transformation results in different distances between points and you have to use a "bendy ruler" to make the distances work, that is exactly what you have done.
agreed again.
The width of Australia should be around 4000km whether calculated in cartesian or celestial coords.
I believe the only discussion we are still having is that i'm claiming the picture below is a sphere expressed in celestial coords, whereas you believe it's a disc expressed in cartesian coords.
Just looking at the width of Australia, we can be pretty sure, cartesian coords are incorrect. When looked at in celestial coords, distances do match with the globe model.
(http://troolon.com/wp-content/uploads/2022/02/cake.png)
Mathematically the above picture is a sphere. Just like the 2 graphs below are both representations of the same sine function yet look totally different. Math only cares about numbers, not shapes.

And this brings us back to my original point: physics works regardless of shape. It only cares the numbers and formulas are correct and surprising enough, the numbers work on different shapes... ergo  the shape of the planet can never be known.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 06, 2022, 08:35:41 PM
You continue to miss the point.  If you did the transformation correctly, the distance would be the same in any coordinate system

Quote
Mathematically the above picture is a sphere. Just like the 2 graphs below are both representations of the same sine function yet look totally different

Just because the coordinates are the same, doesn't mean that mathematically it is a sphere.  A sphere has the least amount of surface area per volume of any shape.  All you've done is manipulate the amount of surface area to fit the same volume.

Check out this page
http://walter.bislins.ch/bloge/index.asp?page=Globe+and+Flat+Earth+Transformations+and+Mappings&q=coordinate+transformation#H_WGS84_Coordinate_System
Example above: Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same. They have different directions and lengths and hence different cartesian coordinates. The lengths of the magenta vectors are shown at |Vglobe| and |Vfe| respectively [/quote]

The vectors represent a measurement of area.  So even though the coordinates are the same, the coordinates represent a different measurement of area. That’s exactly what is not supposed to happen when you transform coordinates from one system to another.  If you transform the metric tensor correctly, the coordinates in any system should represent the same amount of area in any other system.

Quote
Note: Coordinate System Transformations (CS Transf) do not change the length and direction of vectors. They only change the values of the vector components to get the same vector in the corresponding coordinate system.


Translation: A correctly performed transformation doesn't change the amount of surface area.

.Sure, you can manipulate the same volume of space of a sphere and plot it on a disc, but you have to ignore the rules about transforming the metric tensor and therefore aren’t represent reality. You can change the shape of anything to look like anything, but if you ignore the rules of physics doing so, it isn’t much of an accomplishment.

Quote
And this brings us back to my original point: physics works regardless of shape.

Not if you follow the rules of physics to determine the shape.

Also, another question.  What mechanism for gravity are you assuming?  If you are sticking with UA, then you should probably be using Rindler coordinates. 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 06, 2022, 09:13:45 PM
You continue to miss the point.
At least one of us does :)
Can you please answer the question i've asked multiple times before. At what point do you believe the transformation breaks?
1. When i'm switching from cartesian to celestial coords,
2. or when i'm drawing latitude on a straight axis instead of a curved axis?

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Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same.
Quote
FYI: The link only contains black/white pictures. I did not find magenta vectors

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Just because the coordinates are the same, doesn't mean that mathematically it is a sphere.  A sphere has the least amount of surface area per volume of any shape.  ...
Even though the geodetic (Lat,Long,Alt) coordinates of the magenta vectors are the same on the Globe and Flat Earth domain, the corresponding vectors are not the same. They have different directions and lengths and hence different cartesian coordinates. The lengths of the magenta vectors are shown at |Vglobe| and |Vfe| respectively
But my vectors are the same...  A point in celestial coords can be expressed in terms of its vectors as:  x*latitude-vector + y*longitude_vector + z*distance-vector.
(note i'm using vector very liberal here as latitude and longitude are a circular and not straight axis. It's probably more logical to express a point as distance along the equator plus distance along the northpole-southpole-greatcircle)
On the flat cylinder, the same point is still expressed as: x*latitude-vector + y*longitude_vector + z*distance-vector.

So as far as the math is concerned, these 2 points are the same. Same values, same vectors

Quote
The vectors represent a measurement of area.  So even though the coordinates are the same, the coordinates represent a different measurement of area. That’s exactly what is not supposed to happen when you transform coordinates from one system to another.  If you transform the metric tensor correctly, the coordinates in any system should represent the same amount of area in any other system.
area is the same as axis and coordinates of any point are the same.
eg: in celestial coords, the sphere with equation lat=[0-π], long=[0-2π], distance<=R  has volume 4πR²
The flat cylinder with equation  lat=[0-π], long=[0-2π], distance<=R  also has volume 4πR²
There is no difference.

Quote
Note: Coordinate System Transformations (CS Transf) do not change the length and direction of vectors. They only change the values of the vector components to get the same vector in the corresponding coordinate system.
Translation: A correctly performed transformation doesn't change the amount of surface area.
Sure, you can manipulate the same volume of space of a sphere and plot it on a disc, but you have to ignore the rules about transforming the metric tensor and therefore aren’t represent reality. You can change the shape of anything to look like anything, but if you ignore the rules of physics doing so, it isn’t much of an accomplishment.
I'm not ignoring any rules. In fact i'm relying on all these rules not to break the physics. If distances or vectors changed, everything would break.
It's just a different representation, but it are celesitial coordinates. You should calculate distances, angles and areas, with to celestial coordinate formulas.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 06, 2022, 09:23:51 PM
From your web page:

"Intuition: Fill space with concentric sphere’s around the origin. For every sphere , transform it into a disc with an an azimuthal projection and finally insert the disc into a cylinder at a height equal to the radius of the sphere.
transforming a sphere"

You are not "transforming a sphere", you are transforming 3space. The transformation includes flattening each sphere, which is not part of the process of coordinate conversion. Because we are only interested in the earth, we can ignore all of the stack of disks except the one with the radius of the earth. You can shorten the statement to "take the earth and transform it into a disc". This is not changing coordinate systems, this is just flattening the disk, just like the maps in the FAQ and with the same distortion.

If you represent the same sphere in different coordinate systems and use the equations for that coordinate system, you will get the same distances, shape, size in all coordinate systems. To convert a sphere in cartessian to spherical, find the instructions here:

https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_12%3A_Vectors_and_the_Geometry_of_Space/12.7%3A_Cylindrical_and_Spherical_Coordinates

In all coordinate systems, you will end up with a sphere where every point is equidistant from the origin. Your AE/FE projection does not do that, so not a sphere.

On a spherical earth, longitude lines below the equator converge, get closer together. On disk earth (FE/AE) they diverge. Case in point, the coasts of Australia. If we take their longitude as the same on RE and FE, on RE the longitude lines are closer than the equator, less distant. On FE, they diverge, the lines are farther apart, more distance. Distance is not preserved, not equivalent. This is what happens when you "straighten the longitude lines". Distance is distorted. The appearance of the AE confirms that Australia is way too wide. We can do the calculations of what the distance would be with converging longitudinal lines and diverging. Only one can match, the other will be falsified.

Still wondering where Sigma Octantus is. Do I understand your reply to be that you can make a graphic of Sigma Octantus light bending however it needs to so that everyone in the southern hemisphere sees it directly south at an angle of inclination equal to their latitude? Perhaps, like the shape of the earth, no one can ever know? Seems like a pretty weird coincidence that the light would bend however it needs to to make Polaris be directly north at an angle of inclination equal to your latitude.

Do you agree that RE geometry explains this with straight light and a reasonable location for Sigma Octantus? That on FE, no one knows where Sigma Octantus is or how or why the light bends?


Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on February 07, 2022, 01:27:27 AM
I don't get what all the fuss is about, really, unless I've missed something.

Quote from: troolon
But as a general answer, my 3D model is globe physics, just rendered differently. If you take a globe, express it in in celestial coords (lat, long, distance), and then draw latitude on a straight instead of a radial axis, you'll get a flat-earth-universe. Physics works with celestial coords, and so it behaves like a globe, it just looks flat. So if a wave works on a globe, it also works on the flat renderering. The only difference is the shape of one axis.
The OP him/herself said that the model is globe earth rendered on a flat disk, with whatever it takes to make the shape of the sun's illumination match the observations, and calling the boundary of the disk a  "mathematical artefact".  It is not flat earth. The title of the thread is a bit disingenuous.

Saying that you can't differentiate between the models is also strange, since there aren't two models, but only one represented in different ways.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 07, 2022, 02:13:30 AM
The OP is claiming that the distances and physics of his model work the same as RE. This is not true, a disk is not equivalent to sphere, the distances are different. What he claimed was changing coordinates included flattening a sphere into a disk. Light has to bend in ways that there is no evidence of. He claims the earth can be any shape, physics works the same, they are mathematically equivalent. This is not true.

You can't prove the earth is round, but you can do 2 things:

1. Assume the earth is flat and explore the implications. You can observe that distances on the AE map are wrong, that there is no good explanation for some people seeing the dome as daylight while at the same time others see it as night. In the northern hemisphere, people see the north star directly north at an inclination equal to your latitude. This works on AE/FE for left/right (azimuth), but there is no place where it works for up/down (altitude) without bending the light. This is called Electromagnetic Acceleration in the FAQ, which the FAQ says is "unknown forces with unknown equations". In the southern hemisphere, everyone can see the southern pole star directly south, which means outward from the globe, thus in Australia and South Africa directly opposite directions at the same time.

This means either the laws of physics are wrong and measurement has to be flexible - he says FE requires bendy rulers, and he is right, but that's not measurement, it is adjusting your observations to fit. This falsifies the model.

2. You can observe that RE explains this quite reasonably with light traveling straight and no contradictions with observations.

The OP made several errors of thought, thinking he was doing a coordinate conversion when he was transforming the object. thinking that the disk he changed it to was a sphere when looking at it it is clearly a flat circle, using spherical formulas on 2d polar coordinate object, thinking that transforming the shape would not change the physics.

RET explains this and much more, OP can't explain where Sigma Octantus is. He can't show day/night sky on the dome, where half the world sees light blue daylight and half see darkness with stars over the entire dome at the same time.

If he wants to say it is a thought experiment illustrating an impossible world, well, that's cute. If he wants to say his model is valid and a possible alternative to RE with equivalent physics, he is wrong.

Again, his conversion flattens the globe, while he claims he is doing a coordinate conversion. It is not the same geometric shape. A sphere is not a disk. You can map the surface of a sphere onto a disk, but that is not the same model.
Title: Re: Found a fully working flat earth model?
Post by: scomato on February 07, 2022, 02:54:23 AM
If OP is claiming that via arbitrary transformation the globe can be a disc and vice versa, what’s to say then the Earth isn’t a rhombus, donut, non-euclidean, or shaped exactly like Howard Stern?

If we ignore reality we can arbitrarily transform one coordinate system into another, welcome to the history of map making, this was never even a point of debate in the first place and I fail to see how it has any relevance to Flat Earth theory at all.
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 07, 2022, 02:17:56 PM
If OP is claiming that via arbitrary transformation the globe can be a disc and vice versa, what’s to say then the Earth isn’t a rhombus, donut, non-euclidean, or shaped exactly like Howard Stern?
If you had read basically anything OP posted in this thread you are posting in right now, you'd know they said exactly this already.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 08, 2022, 10:12:34 PM
...
Because we are only interested in the earth, we can ignore all of the stack of disks except the one with the radius of the earth. You can shorten the statement to "take the earth and transform it into a disc". This is not changing coordinate systems, this is just flattening the disk, just like the maps in the FAQ and with the same distortion.
Why would a flattening not be a coordinate transformation? You can go back and forth between them, no information loss. The surface of a sphere is a 2D structure, so it a disc. Here's the transform to go back and forth between both:
(lat, lon, dist) = (asin(z / √(x²+y²+z²)), atan2(y, x), √(x²+y²+z²))
(x,y,z) = (dist * cos(lat) * cos(lon), dist * cos(lat) * sin(lon), dist *sin(lat))

Quote
If you represent the same sphere in different coordinate systems and use the equations for that coordinate system, you will get the same distances, shape, size in all coordinate systems. To convert a sphere in cartessian to spherical, find the instructions here:
In all coordinate systems, you will end up with a sphere where every point is equidistant from the origin. Your AE/FE projection does not do that, so not a sphere.
Please remember that on a sphere in spherical coordinates, coordinates are expressed as (lat, lon). Or degrees along the equator-circle+ degrees along the NS-circle. In the flat rendering points are still expressed as degrees along the equator-circle+ degrees along the NS-line
I know this sounds counterintuitive as we use cartesian coordinates so often, but on the cylinder, the 3 axis are the equator, NS-line and height. Any point is expressed as a "sum of these 3 axis".  When using these 3 axis, all points are indeed equidistant from the center of the earth.
The easiest way to show is:   center of the earth in spherical coords = (0,0,0). Any point on the flat earth has coordinates([0-π], [0-2π], 6000km). The distance between any point and the origin is exactly 6000km. Exactly as expected.
Do note that if we represent coordinates in cartesian form (x,y,z) and then calculate the distance, everything will be broken. You must respect the axis.

Quote
On a spherical earth, longitude lines below the equator converge, get closer together. On disk earth (FE/AE) they diverge. Case in point, the coasts of Australia. If we take their longitude as the same on RE and FE, on RE the longitude lines are closer than the equator, less distant. On FE, they diverge, the lines are farther apart, more distance. Distance is not preserved, not equivalent. This is what happens when you "straighten the longitude lines". Distance is distorted. The appearance of the AE confirms that Australia is way too wide. We can do the calculations of what the distance would be with converging longitudinal lines and diverging. Only one can match, the other will be falsified.
Remember our axis are the "equator" and NS-line. On a sphere, points are expressed as (lat/lon) -- degrees along the equator + degrees on the NS-circle. On the flat earth they are too. When you express coords as lat/lon, you should use the lat/lon distance formulas (ie haversine) and Australia will have the correct width also on the flat earth.
To give a quick example. the left-most point of australia on the flat map has coord (-33°, 115°). The rightmost point (-29°, 154°). Calculating the distance gives: 3722km (which matches google maps and the globe model)
Calculating the distance with xy-coordinates will indeed give horribly deformed results.

Quote
Still wondering where Sigma Octantus is. Do I understand your reply to be that you can make a graphic of Sigma Octantus light bending however it needs to so that everyone in the southern hemisphere sees it directly south at an angle of inclination equal to their latitude? Perhaps, like the shape of the earth, no one can ever know? Seems like a pretty weird coincidence that the light would bend however it needs to to make Polaris be directly north at an angle of inclination equal to your latitude.
If you had asked for 3 people looking north and seeing polaris, the answer would be really straightforward on the disc. Draw a star above the northpole, draw curvy rays from the 3 observers going to the star and done.
To understand the southstar, there's one peculiarity about the AE map you must understand: The southpole, a single point on the globe, gets transformed into a circle. The entire outer edge of the AE map represents just 1 point in reality. Mathematically: the southpole has coordinate (lat,lon) = (-90°, [0-2π]) There is no unique coordinate for this point.
The same phenomenon happens with points directly below the southpole: it gets transformed into a circle. So the southstar in the flat earth universe is rendered as a circle rather than a point. It's rather obvious that 3 people looking south, can see the same circle.
If we move the southstar just the slightest bit away from the southpole, it would again become a single point. When you then draw the rays of light, you will see they start due south from the observer, and then start curving all around the disc to meet up with the southstar. I can promise you, mathematically it is all correct even though it's not very intuitive.

Quote
Do you agree that RE geometry explains this with straight light and a reasonable location for Sigma Octantus? That on FE, no one knows where Sigma Octantus is or how or why the light bends?
Re explains the world very well.
On FE, sigma octantis position is given by (lat, lon, distance). We know perfectly well where it is.
The reason light bends in the FE rendering is the same reason light travels straight in RE.
Remember both RE and FE share the same physics. There's not difference. It's just a different representation.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 08, 2022, 10:22:59 PM
I don't get what all the fuss is about, really, unless I've missed something.
Based on your reactin, you have probably totally understood.
From a physics viewpoint this model has nothing new to offer.
However in the context of the flat-earth-"debate", it offers a few interesting tidbits:
- There exists a working flat-earth model. People have been looking for this for over 100 years
- The true shape of the earth can never be known. It can be a globe, or flat or a pyramid. There's no way to measure it. It seems people do not always realize this.

But yes, you have probably understood everything correctly and your reaction is very typical for a physics graduate.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 08, 2022, 10:29:23 PM
If OP is claiming that via arbitrary transformation the globe can be a disc and vice versa, what’s to say then the Earth isn’t a rhombus, donut, non-euclidean, or shaped exactly like Howard Stern?
I totally agree. Physics can be made to work on any shape planet.

Quote
If we ignore reality we can arbitrarily transform one coordinate system into another, welcome to the history of map making, this was never even a point of debate in the first place and I fail to see how it has any relevance to Flat Earth theory at all.
We're not ignoring reality. Physics works regardless of shape. You can make physics work equally well on a globe, a disc or Howard Stern.
However as all these shapes produce the same answer to any calculation, it's impossible to differentiate the models.
And so in reality there's no way to measure if we're on a globe, or a disc. We could be on a globe with straight light, or we could be on a disc with bendy light.
There's no measurement or observation to find out.
And in my mind, this pretty much settles the entire flat-earth-"debate". Any shape earth is possible and we'll never be able to find out what the real shape is...
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 08, 2022, 10:58:01 PM
The OP is claiming that the distances and physics of his model work the same as RE.
<paraphrased> but nothing matches </paraphrased>
I've given you the math.
I've given you examples.
I've told you how you should calculate distances.
I've created loads of images and animations: day, night, seasons, southstar, ...
I've explained how to create the images yourself.

The graphics and animations match expectations in reality (ie day/night/seasons/keels of ships disappearing, ...)
Quite a few people in this forum have understood.
I've had this work validated by physicists prior to posting.
I've managed to explain a computer to do all of this.
I'm currently working on adding all planets in the solar system at their correct positions.
And yet you keep coming back to the same questions that if true would break all of the above results.
I'm not sure anymore how i can explain it any better.

This model is globe physics. Mathematically there is no difference. Yes it looks flat, but to mathematics it behaves as a globe.
Mathematics doesn't care about shape. It can't see what you've drawn on your piece of paper.
The reason it fails your intuition is because there is no XYZ axis. Coordinates are expressed as lat/long: degrees along the equator + degrees along the NS-line.
In this axis everything makes sense.
Title: Re: Found a fully working flat earth model?
Post by: stack on February 08, 2022, 11:27:17 PM
Troolon, you might find this site interesting. It allows you to custom generate a couple of dozen different map projections. I'm not sure what transformations it uses, but it may be the same as yours.

WORLDMAP­GENERATOR (https://www.worldmapgenerator.com/en/)

(https://i.imgur.com/uayzBfT.gif)
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 09, 2022, 07:09:08 AM
You broke physics (actually geometry is probably more accurate) when you decided to just make up your own rules for a metric. I’ve wracked my brain trying to figure out a way to explain to you, in a way that you will understand. It is difficult because you don’t seem to grasp the very basic concept that Euclidean (flat) and non-Euclidean geometry (spherical) do not follow the same rules.  Most importantly, distances are measured differently

In differential geometry, distances are measured by a metric.  Euclidean and non-Euclidean spaces use different metrics. They have to because the definition of  “the shortest distance between two points is different”.  For Euclidean space it is a straight line, for non-Euclidean it is a curve. For Euclidean space the metric is simply the Pythagorean theorem.  In non-Euclidean space it is
(https://i.imgur.com/nfFqWNh.png)
Obviously, very different formulas

When you do a coordinate transformation between two Euclidean spaces, the metric remains the same so lengths and angles are preserved.  When you do a coordinate transformation from a Euclidean space to a non-Euclidean space (Cartesian on a flat disc to Celestial on a sphere, for example), you aren’t just transforming the coordinate system, you are also transforming the metric from Euclidean to non-Euclidean. You seem to have just arbitrarily transformed the metric by “updating the formula”.  You can’t do that. The metric transforms when you do the coordinate transformation. You can’t just decide that you don’t like the way it transforms and invent your own metric.

But even when done correctly, because the bases of the two different metrics (the underlying geometry) are different, it will never transform exactly. Non-Euclidean metric defines distances in terms of angles, Euclidean doesn’t.
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on February 09, 2022, 03:21:15 PM
I don't get what all the fuss is about, really, unless I've missed something.
Based on your reactin, you have probably totally understood.
From a physics viewpoint this model has nothing new to offer.
However in the context of the flat-earth-"debate", it offers a few interesting tidbits:
- There exists a working flat-earth model. People have been looking for this for over 100 years
- The true shape of the earth can never be known. It can be a globe, or flat or a pyramid. There's no way to measure it. It seems people do not always realize this.

But yes, you have probably understood everything correctly and your reaction is very typical for a physics graduate.

First bold quote contradicts what you said. You don't have a flat earth model, you have a globe earth model represented in a disk. A projection, nothing new. Then you make everything else fit to that shape, starting from a globe.

What people are trying to say is that reality gets in the way. If your model is meant to represent a truly, real, physical flat earth, there is absolutely no reason whatsoever for the scale of the map to be different depending on the location, and the distance metric should be euclidian everywhere. This clearly does not work, as you have mentioned about Australia.

On the second point, it can only be right if you disregard physics completely and only care about a purely mathematical model. The shape of sunlight, sunsets and sunrises, constellation retaining shapes throughout the night... all these are observations that point to a globe. You could MAKE them fit any projection of the earth you want, sure. But the instant you claim that any projection is actually reality, you'll need to invoke completely new and unsupported (by evidence) physics to explain the behaviour of light, the fact that people circumnavigate Antarctica, observations of earth from space, etc.

And to finish it, if you calculate the curvature of your model to be different from 0, you don't have a flat earth. I hope this is clear
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 09, 2022, 07:38:48 PM
You broke physics (actually geometry is probably more accurate) when you decided to just make up your own rules for a metric. ...
I'm not making up any rules. In cartesian coords we have 2 distance metrics:
- the pythogorean one (through the earth)
- arclength along a greatcircle.
We will be focusing on the second one as we want to measure distances along the globe.
Next we transform to celestial coords (latitude, longitude).
The distance metric also transforms. It now becomes the haversine formula.
How is this breaking physics or geometry? Isn't this the way any physicist handles coordinate transforms?

I've asked you for several posts now where in my coord transform i'm making any mistakes and i've still not received a reply.
So lets try again. Example will be to calculate the width of Australia.
1. start with a globe in cartesian coords. Distance formula is arclength along a greatcircle
2. convert to celestial coords. Distance formula becomes haversine. Width of Australia is still correct
3. Render latitude on a straight rather than a curvy axis. All coordinates and formulas stay the same. The width of Australia is still correct as mathematically nothing changed.

Please tell me what i'm doing wrong.

Quote
In differential geometry, distances are measured by a metric.  Euclidean and non-Euclidean spaces use different metrics. They have to because the definition of  “the shortest distance between two points is different”.  For Euclidean space it is a straight line, for non-Euclidean it is a curve. For Euclidean space the metric is simply the Pythagorean theorem.  In non-Euclidean space it is
(https://i.imgur.com/nfFqWNh.png)
Obviously, very different formulas
I don't agree pythagorean distances are the only distance metric. In our discussion we're more interested in arclength along a greatcircle distances.
But fine, let's go with straight distances through the earth.

Quote
When you do a coordinate transformation between two Euclidean spaces, the metric remains the same so lengths and angles are preserved.  When you do a coordinate transformation from a Euclidean space to a non-Euclidean space (Cartesian on a flat disc to Celestial on a sphere, for example), you aren’t just transforming the coordinate system, you are also transforming the metric from Euclidean to non-Euclidean. You seem to have just arbitrarily transformed the metric by “updating the formula”.  You can’t do that. The metric transforms when you do the coordinate transformation. You can’t just decide that you don’t like the way it transforms and invent your own metric.
I believe i'm doing the distance transform correct. Here's how i do it, please tell me where i go wrong:
Imagine you have coordinates (x,y,z) and a distance metric d(p1, p2).  (say pythagorean)
Take a reversible coordinate transformation f(), and it's inverse f_1().
Then the transformed coordinates (x', y', z') are defined as  f(x,y,z).  Or inversely:  (x,y,z) = f_1(x', y', z').
We define the distance metric d' as
d'(p1', p2') = d(f_1(p1'), f_1(p2'))
So to find the distance between 2 points in the transformed space, we go back to the original space, and calculate the distance there.
This by definition give the same distance between transformed and original space. Where am i going wrong?
These are not arbitrary formulas. This is the way you're supposed to do coord transforms.

Quote
But even when done correctly, because the bases of the two different metrics (the underlying geometry) are different, it will never transform exactly. Non-Euclidean metric defines distances in terms of angles, Euclidean doesn’t.
So haversine is wrong, and physics should immediately stop using celestial coordinates?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 09, 2022, 07:56:09 PM
First bold quote contradicts what you said. You don't have a flat earth model, you have a globe earth model represented in a disk. A projection, nothing new. Then you make everything else fit to that shape, starting from a globe.
All formulas presented here can be derived without relying on a globe. If Copernicus hadn't lived, this model would be the physics today.

Quote
What people are trying to say is that reality gets in the way.
How will you tell?
Both models are indistinguishable. You called them a single model.
There is no measurement or observation of reality that can tell them apart. They both represent is equally well.

Quote
If your model is meant to represent a truly, real, physical flat earth, there is absolutely no reason whatsoever for the scale of the map to be different depending on the location, and the distance metric should be euclidian everywhere. This clearly does not work, as you have mentioned about Australia.
Only if you're assuming an orthonormal basis. In an orthonormal basis Australia is broken.
The flat world i'm presenting has the equator and the NS-line as axis and coordinates are expressed in degrees (lat/long along these axis, just like a celestial coordinates)

Quote
On the second point, it can only be right if you disregard physics completely and only care about a purely mathematical model. The shape of sunlight, sunsets and sunrises, constellation retaining shapes throughout the night... all these are observations that point to a globe. You could MAKE them fit any projection of the earth you want, sure. But the instant you claim that any projection is actually reality, you'll need to invoke completely new and unsupported (by evidence) physics to explain the behaviour of light, the fact that people circumnavigate Antarctica, observations of earth from space, etc.
Physics is a model of reality.
We now have 2 models: one with a flat earth, and one with a globe.
How do we know which one is correct? We device a test and check which model matches the results.
Only problem, both models are identical and always give the same result. There's no way to tell which one is right.
Ergo, we can't measure the true shape of the earth.

As for inventing physics: please tell me why light travels straight in a globe world?
My guess is because it matches observations. Same thing with curvy light on a flat planet.
The two models are just one. The explanation remains the same.

Quote
And to finish it, if you calculate the curvature of your model to be different from 0, you don't have a flat earth. I hope this is clear
Only in an orthonormal reference frame. Shapes only make sense if you have defined a basis.

We have presented two differently shaped models of the same physics.
They can't be differentiated by any test.
Therefore it's impossible to know the shape of the earth.

Most likely you might even agree the universe could be a simulation. How would you tell if it's a simulation of a flat world with bendy light or of a globe with straight light? There exists no test.
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on February 10, 2022, 05:40:17 PM
Quote
Quote
And to finish it, if you calculate the curvature of your model to be different from 0, you don't have a flat earth. I hope this is clear
Only in an orthonormal reference frame. Shapes only make sense if you have defined a basis.

No, curvature does not depend on the choice of coordinates. You could define curvature without ever mentioning coordinates.

Quote
We have presented two differently shaped models of the same physics.
They can't be differentiated by any test.
Therefore it's impossible to know the shape of the earth.
You don't have two different shapes. You said it yourself - take the globe, change coordinates, this is what you get. This cannot alter the "shape" you were using, no matter what coordinates you try to use. If your curvature was non-zero to start with, you can't make it zero (globally) just by a change in coordinate system.

Quote
We now have 2 models: one with a flat earth, and one with a globe.
No, still only a globe represented differently than usual
Quote
How do we know which one is correct? We device a test and check which model matches the results.
Only problem, both models are identical and always give the same result. There's no way to tell which one is right.
Ergo, we can't measure the true shape of the earth.
They give you the same result because they are the same model. Try to explain the sun/moon going below the horizon for an observer, but for others is still visible high in the sky. Or the fact that constellations keep their shape throughout the night, or the sun keeping the same size all day, or earth being seen as a globe from space. Of course you could chalk it up to "earth is flat, therefore some weird thing happens". But then you're not devising a test to know the shape, you're assuming the shape and then fitting everything else to that. Or, again, you invoke some handwavy-not-supported physics for that.
Quote
As for inventing physics: please tell me why light travels straight in a globe world?
My guess is because it matches observations. Same thing with curvy light on a flat planet.
The two models are just one. The explanation remains the same.
Light travels straight provided there are no ways to refract, deflect or interact with light. We see they go straight (as in follow a geodesic) because this is what is observed and measured, no matter what the shape of the earth is. The sun is not the only source of light we have, you know. In this case, theory and experiment agree. However, bendy light on a flat earth is a complete ad hoc hypothesis, as it starts with the earth being flat and then it has to explain sunsets, sunrises, etc.
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Only if you're assuming an orthonormal basis. In an orthonormal basis Australia is broken.
The flat world i'm presenting has the equator and the NS-line as axis and coordinates are expressed in degrees (lat/long along these axis, just like a celestial coordinates)
I'm not assuming any basis. If the earth is flat, any map of it is a projection of a disk onto a smaller disk; no distortion happens. That means that if I measured the width of Australia to be 3000km (random number, doesn't matter for the argument) and 3000 pixels in a map, the scale is 1km/pixel everywhere. Therefore, measuring 3000 pixels anywhere else on the map means that distance is also 3000km.
And just to reiterate, distance is invariant by a change in coordinate system.

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How will you tell?
Both models are indistinguishable. You called them a single model.
There is no measurement or observation of reality that can tell them apart. They both represent is equally well.
They are a single model, and it was not me who said it, it was you. You took a globe earth, changed coordinates and projected it in a disk. Everything else was made to fit that disk based on the observations made on a globe earth.



Title: Re: Found a fully working flat earth model?
Post by: Rog on February 11, 2022, 04:40:52 AM
Quote
I don't agree pythagorean distances are the only distance metric. In our discussion we're more interested in arclength along a greatcircle distances.
But fine, let's go with straight distances through the earth.
It isn’t the only distance metric, that’s the whole point.  But there are defined metrics for Euclidean and non-Euclidean spaces.  You can’t just make up your own based on how you want it to turn out and expect your model to reflect reality.
(https://i.imgur.com/jtq2ReO.png)
That’s the metric for non-Euclidean spaces. It takes arclengths, angles and great circles into account.
I can’t help you with the math, but if you aren’t using that formula for your metric, your model doesn’t reflect reality.
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These are not arbitrary formulas. This is the way you're supposed to do coord transforms
Some coordinate systems are inherently Euclidean (they only have two coordinates) and some non-Euclidean (three coordinates) The metric is baked into whatever coordinate system you are using.  If you transform from a spherical coordinate system to another spherical coordinate system, there is no distortion because they both have the same metric. When you transform from a coordinate system that is inherently Euclidean to one that is non-Euclidean, the transformed system becomes non-Euclidean and there is distortion.

If you are measuring the triangle on a sphere you get the measurements on the right, if measuring on flat space, you get the measurements on the left.
(https://i.imgur.com/WxPfaO7.png)
Distortions are the result of using a non-Euclidean metric in a Euclidean space. If there is a triangle that in reality, on a sphere earth, looks like the one on the right, it will look like the one on the left on a flat space. It won’t be accurate.  Your model is inherently distorted because you are using celestial coordinates, which have three coordinates,  and projecting them onto a Euclidean space, which is measured in only two coordinates.
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The geometric properties of the space depends on the metric chosen, and by using a different metric we can construct interesting non-Euclidean geometries such as those used in the theory of general relativity.
http://wiki.gis.com/wiki/index.php/Metric_space
You can apply any random metric to any shape, using any coordinate system and mathematically “change” the underlying geometry but if you are mixing and matching Euclidean and non-Euclidean spaces and metrics, your are always going to have distortion from an actual real, physical structure.
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How will you tell?
Both models are indistinguishable. You called them a single model.
There is no measurement or observation of reality that can tell them apart. They both represent is equally well.
Your logic is backwards.  You are starting with the assumption that the earth doesn’t have real, physically defined geometry to begin with and the geometry isn’t defined until some arbitrary, random metric is applied to it. We can't know the correct geometry because we don't know the "correct" metric.
.
But we do know the correct geometry of the earth. It has intrinsic curvature, which by definition, can be measured by the “inhabitants”.  You keep saying that there is no test or observation we can use, but that is just wrong.  There are lots of them.
(https://i.imgur.com/is6YnaF.png)
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10.2  Parallel Transport in a Curved Space, D=2
Consider the scenario shown in figure 12. It starts out exactly the same as the previous scenario. However, in this case we suppose that the arrows exist in a two-dimensional space, namely a sphere, i.e. the surface of the earth.

The yellow arrows along the equator are exactly the same here as in the previous scenario. Even though the arrows are the same, we have to describe them differently. We say they all point north along the earth’s surface. They point toward the earth’s geographic pole, not toward the celestial north pole, because the latter does not exist in the two-dimensional space we are using.
As we move northward along the leg of the triangle that goes through North America, the arrows in figure 11 continue to point north toward the geographic north pole. Relative to the arrows in figure 12, these arrows must pitch down so that they remain within the two-dimensional space. They are confined to be everywhere tangent to the surface of the earth. As we move north, each of the arrows is parallel to the previous arrow, as parallel as it possibly could be.
Let’s be clear: Each new arrow is constructed to be parallel to the previous one, as parallel as it possibly could be. What we mean by “parallel” is discussed in more detail in section 10.3.
After we get to the north pole, we start moving south along a the prime meridian. We move south through Greenwich and keep going until we reach the equator at a point in the Gulf of Guinea. As always, each newly constructed vector is parallel to the previous arrow. All the arrows on this leg point due east.
Finally, we move west along the equator until we reach the starting point. Again each arrow is parallel to the previous one. All the red arrows on this leg point due east.
At this point we see something remarkable: The final arrow is not parallel to the arrow we started with.
From this we learn that in a curved space, there cannot be any global notion of A parallel to B. We must instead settle for a notion of parallel transport along a specified path. That is: the notion of parallelism is path-dependent. It also depends on whether you go around the path clockwise or counter-clockwise.
If you start with a northward-pointing vector in Brazil and parallel-transport it to the Gulf of Guinea, you get a northward-pointing vector. If you start with the same vector and transport it clockwise around two legs of the triangle as shown in figure 12, you get an eastward-pointing vector.

Creatures who live in the curved space can perceive this in a number of ways. Careful surveying is one way. Gyroscopes provide another way. That is, a gyroscope that is carried all the way around a loop will precess relative to a gyroscope that remains at the starting point.
https://www.av8n.com/physics/geodesics.htm

Parallel transport in a flat environment is not path dependent.  It would be a simple matter to test and if it is found that parallel transport is path dependent, then we know we live in a curved environment.  If we want to accurately model it, we know to use a 3 coordinate system, which will give us the correct metric.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 11, 2022, 01:18:16 PM
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I've asked you for several posts now where in my coord transform i'm making any mistakes and i've still not received a reply.
So lets try again. Example will be to calculate the width of Australia.
1. start with a globe in cartesian coords. Distance formula is arclength along a greatcircle
2. convert to celestial coords. Distance formula becomes haversine. Width of Australia is still correct
3. Render latitude on a straight rather than a curvy axis. All coordinates and formulas stay the same. The width of Australia is still correct as mathematically nothing changed.

When you transform to celestial coordinates the metric does not become haversine.. It becomes the formula I posted earlier.  Haversine calculates the shortest distance between two points on a sphere.  It is only accurate for defined areas of a sphere because it only takes the radius of the earth into account between the two points in question.  It only transforms the area between those two points.  The spherical metric transforms the whole sphere.

I was thinking about this last night and maybe this will give you some clarity.

A Euclidean coordinate system has two dimensions, length and width.  Non-Euclidean has three, length, width and depth.   There is no way to visually depict depth on a flat surface. So when you transform from 2 coordinates to 3  (or the other way), you are adding or removing a dimension that in reality, doesn’t exist. To account for this extra dimension, shapes or distances are necessarily distorted.

Think about what happens when you break down a cardboard box. In a sense, you are physically transforming from a 3 dimension coordinate system to a 2 dimension one.  When you break it down, the box becomes flat, but it also becomes longer and/or wider. The area that formed the depth has to go somewhere.

You can’t break down a sphere like you can a box, but try this.  Take some clay, form it into a sphere. Draw a # on it so it covers the whole surface of the sphere.  Then use a rolling pin or something to flatten it out so that the # covers the whole surface. The # will distort as the sphere becomes more and more flat. That’s because (at least one reason) a sphere has less surface area per volume than a flat surface.  In order to still cover the whole surface, the # has to “spread out”.

Or you can do it the other way and draw the # on flat clay and then form it into a sphere.  The # becomes curvy.  That’s what happens when you transform from a 2 coordinate system to a 3 coordinate system...the metric changes and straight lines become curvy.  There is less surface area so the lines have to curve to still fit into the same amount space.

You could use the haversine formula to calculate the shortest distance between two points, but the # is still curvy. Just arbitrarily drawing new lines after the fact doesn’t change the fact that the lines had curve in order to still fit into a smaller amount of surface area.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 11, 2022, 10:27:36 PM
My understanding of Euclidean is the regular understanding of space (usually 3 dimensions, can be more or less) the thing Descartes graphed. It can be described in any coordinate system, cartessian, spherical, cylindrical, I think pother obscure systems. Non-euclidean is when you start with something that seems to make no sense in the daily world, such as more than one line passes through the same two points, or parallel lines can meet. Everything we are discussing here is euclidean, if you make it complicated it makes it easier to slip wrong stuff in.

I am going to use the physics convention notation, math is different, and there are multiple celestial versions, but it is all equivalent, just coordinates in different orders.

A basis for a space is a set of vectors such that there exists a real number that you can multiply each vector by, add them up, and reach any point in that space. In cartessian coordinates a basis for 3 space is a vector of length 1 starting at 0,0,0 on each of the x, y, and z axis. So we can reach the point 2,3,4 by multiplying the x vector by 2, y by 3, and z by 4. The same thing can be done with spherical coordinates, the three vectors are (1, 90 degrees, 0 degrees), (1, 90 degrees, 90 degrees), and (1, 0 degrees, 0 degrees). The exact same vectors, just shown different coordinate systems. This diagram shows both the angles of spherical and a cartessian background. so you can see these are two measurement systems overlaying the same physical reality. Not sure what this has to do with the physics we are discussing or the shape of the earth.

(https://upload.wikimedia.org/wikipedia/commons/thumb/4/4f/3D_Spherical.svg/360px-3D_Spherical.svg.png)

Now consider a sphere of radius 3. In cartessian, the formula for a sphere is radius 3 is 3 = sqrt(x2 + y2 + z2). The same sphere in spherical is r = 3. If you graph this on the above diagram, you can use the caretessian to diagram it using the squares in the background. We can diagram the exact same sphere with spherical, easy as all points 3 units away from 0,0,0 Same sphere.

I do not know what Troolon means by coordinate conversion, but what I understand is that coordinate systems are measurement grids superimposed over the same reality. Converting coordinate systems means describing the object using different equations to get the same object using a different grid.

The disk (FE/AE) that Troolon says happens when you convert coordinates is not a sphere, the definition of a sphere is every point on the surface is equidistant to the center, not true of disk/AE/FE. The equation of a disk in the x,y plane of cartessian coordinates is r <= x2 + y2. You don't need z because a disk is 2d. In spherical, ii is disk = r <= r, rho = 90, for all theta. Again, only need 2 coordinates because 2d figure.

The disk has different equations, different shape, and disk does not meet the mathematical definition of a sphere. Coordinate systems do not change or control reality, they are different ways of describing the same thing.
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 11, 2022, 10:45:21 PM
Troolon,

I am curious where you learned math. Did you take classes in trigonometry/solid geometry and linear algebra? Perhaps you are self taught by looking at web pages about this stuff in the curse of your FE research? Did you figure it out for yourself? Do you have any reason to believe you understand it correctly other than your own self confidence, like a test, degree, certificate, or ??? Is there someone more certain to correctly understand math, or are you the ultimate authority.

I would like to go with you to see a math teacher and present your ideas on coordinate conversion turning a sphere into a disk.

It is hard to understand what you are saying when my understanding of the basic ideas is so different than mine, but as I understand you, your idea is that if something is mathematically equivalent then the physics is equivalent. Since you can convert anything to anything per your coordinate conversion technique, the physics of all shapes is the same.

I would like to go with you to a physicist and present your theory that the physics of all objects are the same.

I would expect the mathematicians and physicists to give you an explanation much ike mine and reject your ideas. If that happened, would you claim that all math and physics teachers were involved in a conspiracy to hide the true shape of the earth by teaching wrong things? Or would you say they are all dumb and only you figured it out correctly?

What percentage of mathematicians and physicists (and astrophysicists and astronomers) would agree with your ideas. My guess is zero.

I saw a video of Neil De Grasse Tyson talking about flat earth, he said it was impossible. Perhaps you understand math and physics better than him? Or he is desperately trying to fool us?


Title: Re: Found a fully working flat earth model?
Post by: jimster on February 11, 2022, 11:08:36 PM
Troolon,

Is this a correct summary of your ideas?

1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original.
3. Flattening a sphere into a disk is a coordinate change.
4. Therefor, a disk is mathematically equivalent to a disk.
5. Because a disk is mathematically equivalent to a sphere, the physics is equivalent.
6. Because it is physically equivalent, light must bend however it needs to to change the RE appearance into FE reality.
7. Because the disk and sphere are mathematically equivalent, the equation for distance on a sphere can be used on a disk.
8. If the distance calculation on a disk does not match observed reality, then the process of measurement needs to be flexible to be made to match, measurement is not being done right.

Did I get it right?
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 12, 2022, 12:01:02 AM
No, curvature does not depend on the choice of coordinates. You could define curvature without ever mentioning coordinates.
Draw a straight line. Now place a logarithmic X and Y axis somewhere. Poof... intrinsically curved line.

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You don't have two different shapes. You said it yourself - take the globe, change coordinates, this is what you get. This cannot alter the "shape" you were using, no matter what coordinates you try to use. If your curvature was non-zero to start with, you can't make it zero (globally) just by a change in coordinate system.
Mathematically the surface is definitely curved.
But to a casual observer it definitely looks flat.
It's just a representation.

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They give you the same result because they are the same model. Try to explain the sun/moon going below the horizon for an observer, but for others is still visible high in the sky. Or the fact that constellations keep their shape throughout the night, or the sun keeping the same size all day, or earth being seen as a globe from space. Of course you could chalk it up to "earth is flat, therefore some weird thing happens". But then you're not devising a test to know the shape, you're assuming the shape and then fitting everything else to that. Or, again, you invoke some handwavy-not-supported physics for that.
I'm not assuming shape. I'm literally saying, we can't know the shape it can be any shape.

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Light travels straight provided there are no ways to refract, deflect or interact with light. We see they go straight (as in follow a geodesic) because this is what is observed and measured, no matter what the shape of the earth is. The sun is not the only source of light we have, you know. In this case, theory and experiment agree. However, bendy light on a flat earth is a complete ad hoc hypothesis, as it starts with the earth being flat and then it has to explain sunsets, sunrises, etc.
Bendy light matches observation.
Shine a laser beam over a lake. You measure curvature. This curvature can be either explained by the earth bending or light bending. Impossible to tell.
There is no test to differentiate the shapes.

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I'm not assuming any basis. If the earth is flat, any map of it is a projection of a disk onto a smaller disk; no distortion happens. That means that if I measured the width of Australia to be 3000km (random number, doesn't matter for the argument) and 3000 pixels in a map, the scale is 1km/pixel everywhere. Therefore, measuring 3000 pixels anywhere else on the map means that distance is also 3000km.
And just to reiterate, distance is invariant by a change in coordinate system.
You're assuming orthonormal basis through all of this paragraph:
- map -> orthonormal basis.
- projection -> orthonormal basis (using curvy projection lines will give you a flat map btw)
- scale -> uniform scales only work for resizing between orthonormal basis....
An observer in reality will measure Australia as 3000km.
The flat-earth model will calculate the width of Australia as 3000km.
Taking a ruler to a model is equivalent with an observer outside of the universe measuring the earth and we can't know what basis or what ruler such a being would use.
If you insist on measuring reality with an orthogonal ruler and drawing it on an orthonormal axis, you will indeed get a sphere. But then you are pre-supposing an orthonormal axis.

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They are a single model, and it was not me who said it, it was you. You took a globe earth, changed coordinates and projected it in a disk. Everything else was made to fit that disk based on the observations made on a globe earth.
Quite right, and yet we now have 2 different representations.
It's like Einstein's:
"am i moving or is the universe moving around me" or
" is earth rotating, or is earth stationary and space rotating around it"
There's no way to tell. It's a matter of representation/perception/reference frames.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 12, 2022, 01:20:49 AM
Hello Rog,

When you take a 2D graph of a line, and transform it to polar coordinates, it will still be a line with the same length and angle as before.
why does this work?
Polar coordinates are non-euclidean: they have curvy axis for one and the coordinates for every point are radically different.

In my transformation i'm doing the exact same thing. (celestial coords are really 3D polar coords)
We draw a line/sphere/plane/... in 3D cartesian space. Then we transform every point to (latitude, longitude, distance), just like polar coords.
The line/sphere/plane is still a line/sphere/plane.  All distances still match with the cartesian drawing.

BTW, I appreciate all the effort you put into your posts, i want you to know that i'm looking into the math until i think i understand most of it.
The only bit i can't say i fully understand yet is the distance formula with the differentials. However i do not wish to debate this formula and will assume it as true. I also don't think the exact formula matters much, any distance metric is fine.

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A Euclidean coordinate system has two dimensions, length and width.  Non-Euclidean has three, length, width and depth.   There is no way to visually depict depth on a flat surface. So when you transform from 2 coordinates to 3  (or the other way), you are adding or removing a dimension that in reality, doesn’t exist. To account for this extra dimension, shapes or distances are necessarily distorted.
In physics and mathematics, the dimension of a mathematical space (or object) is informally defined as the minimum number of coordinates needed to specify any point within it. -- https://en.wikipedia.org/wiki/Dimension
The surface of a sphere (a shell) is a 2D. You only need 2 coordinates to describe any point (latitude, longitude). A flat disc is also 2D. Both implicitly assume R=6000km.
The sphere itself is 3D, and this transforms to a 3D cylinder. (no assumption of R)

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When you transform to celestial coordinates the metric does not become haversine.. It becomes the formula I posted earlier.  Haversine calculates the shortest distance between two points on a sphere.  It is only accurate for defined areas of a sphere because it only takes the radius of the earth into account between the two points in question.  It only transforms the area between those two points.  The spherical metric transforms the whole sphere.
Understood. It's the distance between 2 points on a sphere (haversine), versus 2 points on 2 spheres (spherical metric)
However i don't think the exact shape of the formula matters much. All i care about is that whatever distance formula you use in cartesian coords, you use one that gives the same answer in celestial coordinates. Just like with the polar coordinates i don't want to break anything.


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I was thinking about this last night and maybe this will give you some clarity.
Think about what happens when you break down a cardboard box. In a sense, you are physically transforming from a 3 dimension coordinate system to a 2 dimension one.  When you break it down, the box becomes flat, but it also becomes longer and/or wider. The area that formed the depth has to go somewhere.
I think that depends on if you're working with the surface of the box (2D) or the volume of the box (3D). I do agree you can't have information loss when transforming.
Also i don't want you lying awake about this :)
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You can’t break down a sphere like you can a box, but try this.  Take some clay, form it into a sphere. Draw a # on it so it covers the whole surface of the sphere.  Then use a rolling pin or something to flatten it out so that the # covers the whole surface. The # will distort as the sphere becomes more and more flat. That’s because (at least one reason) a sphere has less surface area per volume than a flat surface.  In order to still cover the whole surface, the # has to “spread out”.
I understand, but that's not what i'm doing.

For a better analogy of what i'm doing, have a look at both plots below.
https://www.wolframalpha.com/input?i=polar+plot+sin+x
https://www.wolframalpha.com/input?i=plot+sin+x
What i believe you may be doing is looking at the polar plot and exclaiming: "that's a circle and not a sine-wave!"

I'm taking some points in cartesian coords, and then expressing them in celestial coords.
That's something i can do without breaking distances or angles imo. -- physics does this all the time.
Then in a second step, i'm representing my polar coords, on straight axis (like the angle on a sine graph is also represented by a straight axis instead of a radial one)
This second step doesn't change formulas or coordinates. It's just a representation change. So this also doesn't break anything.
However after this second step, you must take care not to look at the drawing as cartesian coords, because then everything will be all wonky.
Title: Re: Found a fully working flat earth model?
Post by: stack on February 12, 2022, 03:14:34 AM
Bendy light matches observation.
Shine a laser beam over a lake. You measure curvature. This curvature can be either explained by the earth bending or light bending. Impossible to tell.
There is no test to differentiate the shapes.

What if you don't use a light/laser to survey? Like maybe a Theodolite.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 12, 2022, 05:59:26 PM
Bendy light matches observation.
Shine a laser beam over a lake. You measure curvature. This curvature can be either explained by the earth bending or light bending. Impossible to tell.
There is no test to differentiate the shapes.

What if you don't use a light/laser to survey? Like maybe a Theodolite.
I'm no expert on theodolites, but i believe they're still sight based.
But even in general, it's mathematically impossible to find a difference as flat and globe are the same model.
Title: Re: Found a fully working flat earth model?
Post by: drand48 on February 12, 2022, 06:46:20 PM
Hello,

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
Excellent!  I'm convinced.  So, everyone, just pick whatever model you like and have fun with it.

Just note that (despite what Trolodon saysthink), you can't make a ruler that works in the flat model.  The length of the ruler has to change depending on your latitude and the direction you're pointing, which ... um ... doesn't actually happen.  (Or alternatively, it does, in a way that makes the Earth seem round even though it's flat.)

Great graphics in any case!
Title: Re: Found a fully working flat earth model?
Post by: DuncanDoenitz on February 12, 2022, 08:36:46 PM
@Troolon; can we go back around 5 pages to our carpet-fitting business? 

It turns out that we are quite the success story, and we have expanded into carpet production and installation; our factory in the West Midlands of England receives orders from far and wide, including from the Lord Mayor of Hobart, Tasmania, who wants to carpet his office.  We send out a technician to measure-up and (conveniently) he finds the office to be exactly 3m x 4m, with diagonals of 5m, so it's precisely a right-angle rectangle.  Our representative makes a drawing of the office floor and e-mails it to Company HQ. 

Question 1; Hobart is around 42 deg S latitude.  At what point, if at all, should our technician be using bendy rulers or variable-scales to draw his map, because the drawing is, after all, a map? 

The federal Australian government gets to hear about the Lord Mayor's carpet and they are well impressed with the quality, precise manufacture to specified dimensions, and its absolute resistance to stretch and shrinkage.  It can be rolled up, but it is incompatible with curvature in more than one dimension. 

In an effort to brighten up Australia, they decide to cover the complete country in a nice floral pattern axminster.  A rectangular remnant precisely 3000km x 4000km should do it, and extend a little over the coast.  So we manufacture a carpet loom 3000km wide, and use it to produce a carpet 4000km long.  We know it is a precise right-angle rectangle, because the loom is precisely 3000km, and with our top-notch quality control (the technology of which which we can leave someone else to work out) we used it to produce a carpet exactly 4000km long, without any warpage whatever.  We ship it off to Canberra.  (Again, we can leave the logistics to FedEx or whatever). 

So now the Aussies roll it out from Brisbane towards the West. 

Question 2; How does it look?  Is it wrinkled at the edges?  Is there a ridge over Melbourne? If we push it down over Victoria, does it leave a gap over Sydney?

Perhaps it lies perfectly flat; no gaps.  We have proven that the Earth is flat, or cylindrical, or a cone, or a pyramid or some-such.  We possibly don't know which, but we have completely refuted the idea that the world is spherical, or shaped like Angela Merkel.  If we continue to identify what shapes it isn't, eventually we will know what shape it is. 

Confession time; unbelievably, the above story is not true.  (We went bust in the first year; the clever money, apparently, is in laminate).  The point though, is that as a thought experiment it is true.  If we did it, either the carpet would lie flat, or it wouldn't.  It can't do both.  It would falsify one part of the argument. 

Your contention that we cannot know the shape of anything is a nonsense.  Its not just about the maths.  The Earth, like our fictional carpet, is an entity with fixed dimensions.  If we measure it enough ways, using a constant metric against entities who's size we know, we can know its shape or, at the very least, identify what shape it isn't. 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 12, 2022, 10:12:44 PM
Hello Jimster,

I've always considered myself pretty good at at math, and i've studied sciences at uni, so i've seen my fair share of math problems.
That being said i've not studied maths or physics, and things like gaussian curvature are new to me and took me some time to read up on.
Even now i won't claim much more than some basic understanding.

I actually stumbled on this coord transform mostly by accident, and so i had it scrutinized by graduated physicists, knowledgeable in the field, prior to making it public.
They called it "logically sound, but practically useless" (meaning that in most cases the math is harder than the globe model so they won't be switching over anytime soon :)
I've also accidentally ended up in a discord full of physicists. To them i only had to say the first 4 lines of my idea they all accepted it, no questions asked.
So yes, i believe it passes scrutiny by people truly knowing what they're doing.

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I would expect the mathematicians and physicists to give you an explanation much like mine and reject your ideas. If that happened, would you claim that all math and physics teachers were involved in a conspiracy to hide the true shape of the earth by teaching wrong things? Or would you say they are all dumb and only you figured it out correctly?
All graduated physicists i've talked to have no problem with the math (do take a graduated physicist, the physics students sometimes seem to have a harder time)
If the people i had approached had said i was wrong, this idea would be in the bin instead of this forum.

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I saw a video of Neil De Grasse Tyson talking about flat earth, he said it was impossible. Perhaps you understand math and physics better than him? Or he is desperately trying to fool us?
I think Neil De Grasse Tyson is a physics communicator. His job is to make physics easy explainable to the masses.
I don't think he would be doing a very good job if he were to address every possible detail that amends the mainstream view.
There are also several advantages to the globe view: the math is generally easier, and it's orthogonal (meaning distances scales linearly)
However what many people now forget is that this is just a representation and that there are others.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 12, 2022, 10:32:41 PM
Excellent!  I'm convinced.  So, everyone, just pick whatever model you like and have fun with it.
The way i interpret Occams razor is to use the model that suits your needs best.
For planting a garden bed or assembling furniture, i would recommend flat-earth physics over the globe model.

Personally i think the main uses for this flat-earth-model are:
- the philosophical implications: that we can't know the true shape of the planet
- possibly for visualizations. For me personally i find a tangible cylinder sometimes more appealing and insightful than infinite 3D space.
I'm currently working on a sidereal versus solar day difference and i think this shows up so much more clearly in the flat-earth representation than in an infinite, unfathomable 3D space (i'll post the animations when they're finished)
I'm also hoping it will bring some balance in the flat-earth debate, where i find both sides are sometimes acting incorrect.

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Great graphics in any case!
Thanks
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 13, 2022, 01:02:07 AM
Occam's razor is not what is most useful, it says the simplest explanation is the best.

For example, if the earth is round, Sigma Octantus would be visible only from southern hemisphere and would appear exactly where it is, at an angle of elevation equal to your latitude, and Polaris would do that in the northern hemisphere, while the light travels straight through a vacuum. All according to known physical laws, with equations that give right answers. Day and night happens consistent with oobservation, startrails, night sky/day sky makes sense.

If the eqarth is flat, you can't say where Sigma Octantus is, Polaris is at the wrong elevation, we know the light bends horizontally and vertically. FAQ says "unkown forces and unknown equations". Your map has Australia way too gig, you fix this by making the length of a meter flexible, thus meaning the speed of light is 50% higher in Australia than US. Bendy rulers with flexible units.

The light bends, the ruler stretches, people see day and night sky over the same dome at the same time due to unknown forces. Not a simple system. I gotta call RE by Occam.

You can start with any set of postulates and develop a mathematical system, there is an infinite number. Yours is non-Euclidean, so Australia is free to be the wrong size yet somehow still the right size, like classic examples of two different lines passing through the same point, or parallel lines crossing. Just curve the light however you need to to make it match oberved, and ignore/explain away the size difference. Is it useful? No, it is misleading. Why not use a diagram of what actually happens in the way it actually works?

As for your notion that we can't be sure about things, that is called solipsism, and has a philosophical history. You can hold the position that nothing can be known except that you exist to ask the question. This notion is useful for FE to dodge inconsistencies, as in "you can never know the distance". I believe that in your system, you can never know how the light bends or where anything is. You need that to explain why RET corresponds to observations if the light is straight, while FET requires unexplained and unquantified light bending.

Gps works by sending a timestamp to your device, where the local time is subtracted, and the speed of light is used to calculate your distance from the satellite, a form of LIDAR. If the light is bending all over the place and we don't know where the satellites are, how does this work? All RET, deniable only with conspiracy, changed laws of physics, and denial of known science.

Can I request that you make another graphic? Pick 10 cities around the globe (more is better) and get their distances with google maps or airline schedules. Take a point at the center of your graphic and plot a point 3963 miles from it (radius of earth). Plot a second city also 3963 from origin and at the end of an arc with center at origin and arc length equal to the distance between each city. Continue plotting cities and their connecting arcs. If the result is a sphere the earth is round. If not, it is not round.

If you get on a plane to Hawaii, do you want the navigation to be by RET or FET?

I don't know who you talked to about your math, but measurement is measurement, the same everywhere or meaningless, and changing coordinate systems in Euclidean (the world we live in) changes nothing about the math, only the notation, not the size or shape.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 13, 2022, 07:01:57 AM
Quote
When you take a 2D graph of a line, and transform it to polar coordinates, it will still be a line with the same length and angle as before.
why does this work?
Polar coordinates are non-euclidean: they have curvy axis for one and the coordinates for every point are radically different.

Polar coordinates are Euclidean.(and 2d)  Euclidean coordinates just mean that distances can be measured using the PT.  That is true of Cartesian and Polar coordinates. I don’t know what you mean by the axis is “curvy”,  (I think you might be confusing Polar coordinates with Spherical coordinates...see my comment below) It transforms correctly because lines are Euclidean and 2d and polar coordinates are Euclidean and 2d.
https://en.wikipedia.org/wiki/Polar_coordinate_system
https://math.stackexchange.com/questions/579721/do-there-exist-any-useful-coordinate-systems-in-euclidean-space-besides-the-cart

But if you use Polar coordinates on a sphere, this is what you end up with.
(https://i.imgur.com/l8kXUJF.png)
The lines are distorted because the PT can’t be used to measure distances on a sphere.

Quote
In my transformation i'm doing the exact same thing. (celestial coords are really 3D polar coords)
We draw a line/sphere/plane/... in 3D cartesian space. Then we transform every point to (latitude, longitude, distance), just like polar coords.
The line/sphere/plane is still a line/sphere/plane.  All distances still match with the cartesian drawing.

On a sphere, it will transform correctly because when you transform the coordinates from Cartesian, the metric also transforms and both systems are 3d.

This is just a guess, but I think you might be confusing Polar coordinates with Spherical Polar Coordinates. Polar coordinates are Euclidean, distances can be measured with the PT. Spherical Polar coordinates are not. They use the spherical metric.  If that’s the case, what you are calling “3d polar coordinates” are Spherical coordinates  (sometimes called Spherical Polar Coordinates)...and as you say they are basically the same thing as celestial coordinates.  https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/Spherical_polar_coordinates.html

But whatever you call them, when you transform from Cartesian onto a sphere, it will transform correctly. Both systems are 3d and the metric changes when you transform the coordinates, so distances are measured with the spherical metric.

The problem comes in when you try and project the transformed sphere onto a flat surface. The metric changes back to the PT, and your distances will be off..  You can do it, but there will be distortion.

Quote
No map from the sphere to the plane can be both conformal and area-preserving. If it were, then it would be a local isometry and would preserve Gaussian curvature. The sphere and the plane have different Gaussian curvatures, so this is impossible
.
https://en.wikipedia.org/wiki/Stereographic_projection

An isometry means that distances and angles are preserved   You seem to be under the impression that a transformation,  when done correctly, never changes distances or angles. That’s not true.
Quote
There are many ways to move two-dimensional figures around a plane, but there are only four types of isometries possible: translation, reflection, rotation, and glide reflection. These transformations are also known as rigid motion. 
https://www.infoplease.com/math-science/mathematics/geometry/geometry-isometries#:~:text=There%20are%20many%20ways%20to,also%20known%20as%20rigid%20motion.

Unless your transformation is one of those types, it’s not going to preserve angles or distances. You can have isometric  transformations between a plane and a plane and a sphere and sphere (or other 3d object, but the types of transformations are more limited), but you can’t have one between and plane and a sphere.

The reason should be obvious.  Angles and distances on a plane and angles and distances on a sphere are different. That’s why they use different metrics. If you preserve them, you aren’t going from sphere to plane, you are just going from one sphere to another sphere.

That’s all I have time for now. I’ll try and respond to more later.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 13, 2022, 09:04:19 AM
Troolon,

Is this a correct summary of your ideas?

1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original.
3. Flattening a sphere into a disk is a coordinate change.
4. Therefor, a disk is mathematically equivalent to a disk.
5. Because a disk is mathematically equivalent to a sphere, the physics is equivalent.
6. Because it is physically equivalent, light must bend however it needs to to change the RE appearance into FE reality.
7. Because the disk and sphere are mathematically equivalent, the equation for distance on a sphere can be used on a disk.
8. If the distance calculation on a disk does not match observed reality, then the process of measurement needs to be flexible to be made to match, measurement is not being done right.
I think that's about right.
However my end-conclusion would be that we can't know the shape of the earth.
I also don't agree with 8: For an observer inside the coordinate system, both systems are indistinguishable. But i'll reply soon to carpeting of Australia post and that will hopefully explain this better.

Last night i think i may have come up with a pretty good analogy. How do you like this:
Imagine there exists a warehouse that can store boxes of apples, pears and bananas.
Imagine your only access to the warehouse is a computer that can show the numbers of different boxes.
Every day, you diligently make a pie-chart of the numbers.
One day a new colleague starts, and he suddenly creates a bar chart.
Discussion immediately ensues: what is the correct shape of chart and what is the true size of the warehouse. Is the warehouse a circle, or a very tall bar?
And obviously we don't have enough information to determine the shape of the warehouse.
Both charts are just a representation. The only way to find the shape of the warehouse is by other means than that computer. (eg: calling a warehouse worker)

It's the same thing with the universe.
Our interface with reality is what we can observe and measure.
Physics is just a mathematical representation of this and it's typically represented as a globe.
However that doesn't mean there can't exist other representations.
I have made a mathematical representation of reality that has a flat earth.
And I've also shown that both representations are equally good.
One isn't better or has more information than the other. Both are equally functional. (by virtue of a coordinate transformation existing between the two)
So the only conclusion we can draw is that both the globe and the flat earth are nothing more than equivalent representations of reality.
We can't know the shape of the universe.

Our interface into the world is what we can observe and measure and that is not sufficient to find the shape.
The only way to know is by a different interface (eg phoning an observer outside of the universe)
Title: Re: Found a fully working flat earth model?
Post by: DuncanDoenitz on February 13, 2022, 09:14:07 AM
Fine, but we live in the warehouse.  We can walk around it, climb it, measure it.  Its a warehouse.
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 13, 2022, 11:21:13 AM
" i had it scrutinized by graduated physicists, knowledgeable in the field, prior to making it public. They called it "logically sound, but practically useless"

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

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

Title: Re: Found a fully working flat earth model?
Post by: troolon on February 13, 2022, 12:03:46 PM
Carpeting Australia
first off: very vivid example. I love you approach. :)

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

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

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

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

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

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

Quote
Your contention that we cannot know the shape of anything is a nonsense.  Its not just about the maths.  The Earth, like our fictional carpet, is an entity with fixed dimensions.  If we measure it enough ways, using a constant metric against entities who's size we know, we can know its shape or, at the very least, identify what shape it isn't.
A sphere is just a representation of reality. A representation in an orthonormal axis.
A cylinder is another representation of reality.
It's like bar-charts versus pie-charts. They're just representations. You can't infer the true shape from a mere representation.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 13, 2022, 12:30:57 PM
Fine, but we live in the warehouse.  We can walk around it, climb it, measure it.  Its a warehouse.
This is why i explicitly stated our only provided interface is the provided computer interface.
But fine, let's explore this new example a little further:

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

In the real world you're always measuring 2 quantities at the same time.
- curvature of the earth versus curvature of light (ie laser of over lake)
- distances on the ground versus shape of your ruler/coordinate systems
- rotation of earth versus rotation of space
- ...
We can only ever make relative measurements.
And mathematically it can be shown that you can remove the curvature of earth whilst changing everything else in the universe to compensate. You wouldn't be able to tell.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 13, 2022, 01:44:20 PM
" i had it scrutinized by graduated physicists, knowledgeable in the field, prior to making it public. They called it "logically sound, but practically useless"

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

Doesn't this suggest that practically, there is, or may be, a difference?
In general the math is more difficult, but it will eventually yield the same result.
Remember how in maths you switch to polar coords because it's sometimes easier to calculate with them?
Quick example: spherical distance in lat/long coordinates is the haversine formula. In cartesian coords it's the double differential Rog showed (or a path-length integral). Result will be the same but haversine is probably easier to work with.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 13, 2022, 02:09:33 PM
Occam's razor is not what is most useful, it says the simplest explanation is the best.
Occams Razor is a philosophical idea, not a law of nature. It does not invalidate any model.

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

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

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

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

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

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

Quote
I don't know who you talked to about your math, but measurement is measurement, the same everywhere or meaningless, and changing coordinate systems in Euclidean (the world we live in) changes nothing about the math, only the notation, not the size or shape.
Then please take a piece of paper and draw a logaritmic X and Y axis. Now please draw a circle. (for example all points at distance 1000 from a center at (1000,1000)).  The circle will be deformed. Have fun measuring that with your regular ruler :)
Title: Re: Found a fully working flat earth model?
Post by: stack on February 14, 2022, 08:57:35 PM
Bendy light matches observation.
Shine a laser beam over a lake. You measure curvature. This curvature can be either explained by the earth bending or light bending. Impossible to tell.
There is no test to differentiate the shapes.

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

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

Precision alignment of the LIGO 4 km arms using dual-frequency differential GPS (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjb-PGlhoD2AhXDH0QIHT-OD_MQFnoECCgQAQ&url=https%3A%2F%2Fdcc.ligo.org%2Fpublic%2F0072%2FP000006%2F000%2FP000006-A.pdf&usg=AOvVaw1kc11ZBjr-kxiw5trWrUnB)

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

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

Using global coordinates, the beam tube centerlines were marked along the foundation slab at points spaced uniformly at ~20 m intervals (the unit length of beam tube sections that were welded together in the field) along both arms.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 14, 2022, 10:53:29 PM
You might find this interesting regarding lasers over distance. Specifically how the 4KM long arms of the two LIGO sites were measured and constructed creating a straight light path in relation to Earth's curvature. There are some transformations they did as well which might be of note.
...
Thanks. You never quite stop to think about all the remarkable feats required for these projects.
In my earlier models, earth was just a cylinder and celestial objects were traveling circles. Initially i even had the sun going backwards :)
In my later ones, i am indeed using WGS84 data to more accurately calculate positions of celestial objects however my earth model is still a cylinder.
Converting the cylinder to WGS84 would create a little bulge around the equator-ring but the entire model would remain perfectly circular. (lat/long goes from [0-2π] regardless of obloidness)
However I'm currently not very keen on making this change as i suspect it will be computationally too expensive and it's probably not that visually interesting.
But thank you for this nugget of information :)
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 15, 2022, 12:00:42 AM
Quote
If you insist on measuring reality with an orthogonal ruler and drawing it on an orthonormal axis, you will indeed get a sphere. But then you are pre-supposing an orthonormal axis.

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

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

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

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

It’s an easy test and you don’t have to assume any shape or even take any measurements. Can you show a parallel transport on you model?
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 15, 2022, 03:44:05 AM
Quote
1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original.
3. Flattening a sphere into a disk is a coordinate change.
4. Therefor, a disk is mathematically equivalent to a disk.
5. Because a disk is mathematically equivalent to a sphere, the physics is equivalent.
6. Because it is physically equivalent, light must bend however it needs to to change the RE appearance into FE reality.
7. Because the disk and sphere are mathematically equivalent, the equation for distance on a sphere can be used on a disk.
8. If the distance calculation on a disk does not match observed reality, then the process of measurement needs to be flexible to be made to match, measurement is not being done right

1. Changing coordinates can turn any shape into any other shape.
2. Because it was only a coordinate shape, the result is equivalent to the original. They are only equivalent if the change was isometric.
3. Flattening a sphere into a disk is a coordinate change. Flattening a sphere is more than just a coordinate change because it also changes the metric and the Riemann curvature tensor
4. Therefor, a disk is mathematically equivalent to a disk. When you transform a disc to a sphere the metric changes.  The Riemann tensor changes from 0 to non-zero.  Therefore, a disk and a sphere are not mathematically equivalent.
Title: Re: Found a fully working flat earth model?
Post by: FLATEARTHPSYOP on February 15, 2022, 06:41:51 PM
Troolon ,  in Blender 3.0 --- would you be able to share me the source code in which I can project t he light paths of a sphere on different geometries that you have shown?
Very compelling

Any guides you can show me on how to do this this, just what you have done? Thanks for your hard wrok
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 15, 2022, 09:33:25 PM
Troolon ,  in Blender 3.0 --- would you be able to share me the source code in which I can project t he light paths of a sphere on different geometries that you have shown?
Very compelling

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

BTW here's my latest graphic: a view of the solar systems showing planet positions from jan 2022 up to dec 2022.
(http://troolon.com/wp-content/uploads/2022/02/solar_system.gif)
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 15, 2022, 10:03:37 PM
BTW here's my latest graphic: a view of the solar systems showing planet positions from jan 2022 up to dec 2022.
(http://troolon.com/wp-content/uploads/2022/02/solar_system.gif)

Looking at this, how is it that the distance from Sydney to LAX is a little over 12,000 km yet the distance from Sydney to Santiago is a little over 11,000 km.
Title: Re: Found a fully working flat earth model?
Post by: DuncanDoenitz on February 15, 2022, 10:22:00 PM
You're using a straight ruler. 

Fully working model, but you need a fully working stretchy-bendy ruler. 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 15, 2022, 10:33:55 PM
Orthonormal coordinates are not the only coordinates in Euclidean space.  Just because the coordinate lines are curved, does not mean that the space is curved.  The coordinate system does not define the whether or not a space is curved.

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

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

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

It’s an easy test and you don’t have to assume any shape or even take any measurements. Can you show a parallel transport on you model?
I think it's explained here: https://youtu.be/YJFTWp31WhQ?t=652 starting at 10:24 till 22:28
    Rieman tensor: Rθφθφ = r²sin²φ
    Ricci scalar R = 2/r²   (r = 6000km)

But to understand this more intuitively i must go again to to the construction. ...
1. We start with a sphere in cartesian coords: (https://upload.wikimedia.org/wikipedia/commons/thumb/7/7d/Parallel_Transport.svg/220px-Parallel_Transport.svg.png)
2. We convert the sphere to geographic coords (lat, lon, dist). The graphic looks the same, angles and distances still measure the same. It still looks like this:  (https://upload.wikimedia.org/wikipedia/commons/thumb/7/7d/Parallel_Transport.svg/220px-Parallel_Transport.svg.png)
3. draw latitude on a straight axis instead of a radial one. This is just a change in representation. We do not change coordinates, distances or angles. There's no change in intrinsic curvature as this step is invisible to the maths. I've attached a graphic of what i think the parallel transport looks like but i'm not 100% certain.
Ultimately this graphic is irrelevant as this last step is just a change in representation (like pie charts vs bar charts and not in coordinates, distances or angles -- at least not intrinsically).

Now if you answer, please tell me which of these two steps is incorrect rather than having me learn up on another field of complicated math :)
Physics switches between cartesian and celestial coordinates all the time. If this breaks angles and distances would break here, a large part of physics would be broken.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 15, 2022, 10:38:50 PM
Looking at this, how is it that the distance from Sydney to LAX is a little over 12,000 km yet the distance from Sydney to Santiago is a little over 11,000 km.
This thing is not in cartesian coordinates (x,y) but in celestial coordinates (latitude,longitude,distance).
When you calculate the distance between places using lat/lon coordinates, you will get the same distances as on a globe. (think haversine-formula)

This graph is not in an orthonormal basis, you can't take a ruler to it and just measure it. Mathematically it checks out though.
Title: Re: Found a fully working flat earth model?
Post by: Dr Van Nostrand on February 15, 2022, 10:59:44 PM
Hey,

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


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

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

Distances can be measured.

What you are really saying is that it is impossible for YOU to know the shape of the Earth. I'm guessing you just don't travel a lot.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 15, 2022, 11:21:32 PM
Looking at this, how is it that the distance from Sydney to LAX is a little over 12,000 km yet the distance from Sydney to Santiago is a little over 11,000 km.
This thing is not in cartesian coordinates (x,y) but in celestial coordinates (latitude,longitude,distance).
When you calculate the distance between places using lat/lon coordinates, you will get the same distances as on a globe. (think haversine-formula)

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

I'm not calculating anything.  I'm looking at your picture.  When you look at the disc, straight line path from Syndney to Santiago goes right past LA and continues for a significant distance.  It's a simple observation.  No math involved. 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 15, 2022, 11:25:16 PM
Distances can be measured.
What you are really saying is that it is impossible for YOU to know the shape of the Earth. I'm guessing you just don't travel a lot.
Problem is my proof is mathematical by nature, so unfortunately  i've proven it's impossible for you to know the shape of the earth too :)
Physics is a model of reality. It's just a representation. You can have different representations of the same reality.
All i've done is constructed a different representation of physics...
So when you drive your car and measure distances that way, both representations will predict the same answer.
Soo.. now that we have different shaped representations of reality that can't be distinguished by any measurement, how will you tell what shape planet you're truly on?
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 16, 2022, 12:55:30 AM
Quote
But to understand this more intuitively i must go again to to the construction. ...
1. We start with a sphere in cartesian coords:
2. We convert the sphere to geographic coords (lat, lon, dist). The graphic looks the same, angles and distances still measure the same. It still looks like this: 
3. draw latitude on a straight axis instead of a radial one. This is just a change in representation. We do not change coordinates, distances or angles. There's no change in intrinsic curvature as this step is invisible to the maths. I've attached a graphic of what i think the parallel transport looks like but i'm not 100% certain.
Ultimately this graphic is irrelevant as this last step is just a change in representation (like pie charts vs bar charts and not in coordinates, distances or angles -- at least not intrinsically).

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

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

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

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

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

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

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

(https://i.imgur.com/VILX68U.png)

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

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

You keep contradicting yourself by stating that your model has intrinsic curvature and also stating that we can’t know the shape of the earth.  The very definition of intrinsic curvature is that “the inhabitants” can know.  So which is it?  Does your model have intrinsic curvature or is it impossible to know the shape of the earth?
Title: Re: Found a fully working flat earth model?
Post by: jimster on February 16, 2022, 01:58:37 AM
Troolon,

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

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

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

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

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

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

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

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

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

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

Where is Sigma Octantus?

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

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

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

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


Title: Re: Found a fully working flat earth model?
Post by: troolon on February 16, 2022, 09:31:59 AM
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.
Title: Re: Found a fully working flat earth model?
Post by: Dr Van Nostrand on February 16, 2022, 12:49:14 PM
Distances can be measured.
What you are really saying is that it is impossible for YOU to know the shape of the Earth. I'm guessing you just don't travel a lot.
Problem is my proof is mathematical by nature, so unfortunately  i've proven it's impossible for you to know the shape of the earth too :)
Physics is a model of reality. It's just a representation. You can have different representations of the same reality.
All i've done is constructed a different representation of physics...
So when you drive your car and measure distances that way, both representations will predict the same answer.
Soo.. now that we have different shaped representations of reality that can't be distinguished by any measurement, how will you tell what shape planet you're truly on?

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?
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 16, 2022, 12:53:44 PM
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?

Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on February 16, 2022, 01:49:33 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'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.
Title: Re: Found a fully working flat earth model?
Post by: DuncanDoenitz on February 16, 2022, 05:28:05 PM
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?
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on February 16, 2022, 05:48:12 PM
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.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 16, 2022, 06:38:34 PM
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.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 16, 2022, 06:42:42 PM
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.
Title: Re: Found a fully working flat earth model?
Post by: Dr Van Nostrand on February 16, 2022, 06:48:51 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?
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 16, 2022, 07:04:36 PM
Have you tried reading the very first page? (https://forum.tfes.org/index.php?topic=19093.msg257575#msg257575) 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 ;)
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 16, 2022, 07:14:52 PM
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. 
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 16, 2022, 07:23:23 PM
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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.

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The problem comes when you try and “flatten” the sphere...without any distortion.
At what point have i tried to "flatten" the sphere?

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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.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously on February 16, 2022, 07:44:09 PM
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.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 16, 2022, 09:07:37 PM
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.

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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.

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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?

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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.

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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.

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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.

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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...

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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.

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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...

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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

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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.

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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.
Title: Re: Found a fully working flat earth model?
Post by: JSS on February 16, 2022, 09:36:51 PM
Quote
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. 

Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 16, 2022, 10:03:32 PM
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.
Title: Re: Found a fully working flat earth model?
Post by: Tumeni on February 16, 2022, 11:36:32 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?
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 17, 2022, 12:13:53 AM
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.
Title: Re: Found a fully working flat earth model?
Post by: Iceman on February 17, 2022, 12:44:00 AM
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
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 17, 2022, 03:55:37 AM
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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?

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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.

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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.

(https://i.imgur.com/MXm8mAX.png)

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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.

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

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

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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.







Title: Re: Found a fully working flat earth model?
Post by: stack on February 17, 2022, 05:51:31 AM
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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.
Title: Re: Found a fully working flat earth model?
Post by: Action80 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.
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff 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
Title: Re: Found a fully working flat earth model?
Post by: JSS 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.

Title: Re: Found a fully working flat earth model?
Post by: Tumeni 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?
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously 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.
Title: Re: Found a fully working flat earth model?
Post by: Action80 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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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....
Title: Re: Found a fully working flat earth model?
Post by: troolon 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...
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously 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?
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 17, 2022, 09:42:41 PM
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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.

(https://i.imgur.com/n7Rbgu3.jpg)

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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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 :)
Title: Re: Found a fully working flat earth model?
Post by: troolon 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.

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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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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.

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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.
Title: Re: Found a fully working flat earth model?
Post by: JSS 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. 

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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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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.
Title: Re: Found a fully working flat earth model?
Post by: troolon on February 18, 2022, 10:42:54 PM
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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.

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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.

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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.

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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.

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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.


Title: Re: Found a fully working flat earth model?
Post by: troolon on February 18, 2022, 10:45:01 PM
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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 :)
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 19, 2022, 09:49:32 AM
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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.

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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

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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

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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

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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

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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?


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 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

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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.

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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/


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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…

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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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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...
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff 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
Title: Re: Found a fully working flat earth model?
Post by: Gonzo 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.
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on February 23, 2022, 01:28:25 AM
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?
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 23, 2022, 02:13:11 AM
Quote
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
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on February 23, 2022, 01:18:36 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).
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog 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.
Title: Re: Found a fully working flat earth model?
Post by: jimster 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.

(https://en.wikipedia.org/wiki/Crux#/media/File:Pole01-eng.svg)

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.

Title: Re: Found a fully working flat earth model?
Post by: troolon 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)
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff 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.

Quote
- 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, ...

Quote
- 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.

Quote
- 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.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 25, 2022, 11:51:46 PM
Quote
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. 
Title: Re: Found a fully working flat earth model?
Post by: troolon 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


Title: Re: Found a fully working flat earth model?
Post by: troolon 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.
Title: Re: Found a fully working flat earth model?
Post by: WTF_Seriously 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.
Title: Re: Found a fully working flat earth model?
Post by: Rog on February 26, 2022, 05:09:00 PM
Quote
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.

(https://i.imgur.com/wp1wUEt.png)

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.

Title: Re: Found a fully working flat earth model?
Post by: ichoosereality 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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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.

Quote
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)

Quote
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.

Quote
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.

Quote
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)

Quote
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.

Quote
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?
Quote
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.
(http://troolon.com/wp-content/uploads/2022/03/constellations.png)
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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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.
Title: Re: Found a fully working flat earth model?
Post by: troolon 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

Quote
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. 
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.
Title: Re: Found a fully working flat earth model?
Post by: ichoosereality 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 (https://thatsmaths.com/2018/12/27/gaussian-curvature-the-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 (https://www.nasa.gov/mission_pages/gpb/)}, 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.
Title: Re: Found a fully working flat earth model?
Post by: Rog on March 03, 2022, 06:21:18 PM
Quote
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.

Title: Re: Found a fully working flat earth model?
Post by: astroman on March 04, 2022, 09:14:45 PM
If you have found a fully working flat Earth model then why are you just talking about in these forums??  You should be getting in touch with Nature magazine cos it will be a truly revolutionary Earth model.  How come no one else has come up with it after all this time? Have you got access to resources, equipment and budgets that no one else has?

It is exactly the sort of thing that Nature magazine is all about. At least it is if it really does work as well as well as you claim it does.
Title: Re: Found a fully working flat earth model?
Post by: drand48 on March 06, 2022, 01:10:50 AM
I have a question for troolon: Does your transform preserve topology?  My guess is that it doesn't.  I would be suspicious of using a transform that doesn't preserve topology in physics, to describe the shape of things in the world.

Topology is the study of the things that are constant regardless of how much you stretch or compress a shape.  According to topology, a donut and a typical coffee mug are identical: they're both a class of shapes known as a "torus."  A sphere is not a torus, and can't be shaped into one without tearing or gluing.  So, when talking about the shape of things, I'm suspicious of any transform that does not preserve topology.

Just because you can define a transform between two things does not mean that they are the same.  They're just mappable.  For example, you can map the positive integers onto the set of all rational numbers, and vice versa.  That means the two sets have the same cardinality (the same number of elements.)  But it doesn't mean they're identical in every way and share all properties.

Likewise, Troolon's transform shows that we can map a spherical Earth and its cosmos into a flat Earth and its cosmos, and vice versa.  But that doesn't mean the two share all properties.  In this case, I suspect the topology differs, though I'm not sure of that.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on March 06, 2022, 10:51:23 AM
I have a question for troolon: Does your transform preserve topology?  My guess is that it doesn't. [...] In this case, I suspect the topology differs, though I'm not sure of that.
The zeroth step here would be forming a hypothesis. Where do you think it differs, and why? An essay on topology is largely redundant if all you have is undefined "suspicions".
Title: Re: Found a fully working flat earth model?
Post by: drand48 on March 06, 2022, 04:00:55 PM
I have a question for troolon: Does your transform preserve topology?  My guess is that it doesn't. [...] In this case, I suspect the topology differs, though I'm not sure of that.
The zeroth step here would be forming a hypothesis. Where do you think it differs, and why? An essay on topology is largely redundant if all you have is undefined "suspicions".
I just asked a question.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on March 06, 2022, 10:34:32 PM
I just asked a question.
OK. The answer is "yes" until you present some evidence to the contrary.

Though, in the future, if you only mean to ask a question, you should limit yourself to just asking it. When more than 95% of your post is irrelevant rambling, you know you've gone way off the mark.
Title: Re: Found a fully working flat earth model?
Post by: jimster on March 08, 2022, 11:00:04 PM
Turning the earth into a different shape and then ignoring the distortion of distance, bending the light rays however they need to be bent, is certainly possible, although no answer for where is Sigme Octantus? Incorporating that into the model produce a rotating bent light tube not consistent with any known physics except the need to explain FET.

A "fully working FE model" would have to include that, and be able to show what you see from the surface of the earth. It would show how when it is sunset in Denver how a person in Salt Lake City sees daylight over the entire dome and someone in St Louis sees dark with stars at the same time over the entire dome from the perspective of someone outside the dome.

A model of RET can make sense from an external perspective. A model of FET has a dome that is simultaneously light blue and black at the same time, with different stars in northern and southern hemisphere. And Sigma Octantus becomes a ring with light tubes that bend the light to the observed elevation and stop you from seeing the circle except for the pint directly south of you.

Rube Goldberg.
Title: Re: Found a fully working flat earth model?
Post by: existoid on April 29, 2022, 05:05:58 PM
@OP

There is a problem with the flat monopole map regarding the necessarily diverging lines of longitude the further south one travels from the equator.

In brief, if longitudes kept getting wider as they go further, then every one of the thousands of WW2 battles, flight missions, and dogfights reported by US, Japanese, British, and Australia airmen and sailors that took place south of the equator must have happened quite differently than all available evidence. The planes that flew to and from carriers (and air bases on islands south of the equator) relied on plotting charts with accurate scales for latitude and longitude, and these plotting charts (of which copies are still extant and can be reviewed) do NOT show diverging longitudes, but longitudes that comport with the generally accepted globe model. (Edit: to clarify: incorrect plotting charts would result in an astonishing and intractably overwhelmingly majority of planes flying south of the equator in WW2 to be lost at sea because they could not find their carrier or air base in the vast waters).

More in depth analysis in my original post on this topic here:
https://forum.tfes.org/index.php?topic=16428.msg212915#msg212915


To all (especially the mods) - I apologize if this seems at first glance off topic, given that OP presented his model as one that can explain all the observations of physics. I argue that this is on topic because it is in reaction to the topic and summarizing sentence of the OP:
"I believe to have found a fully working flat earth model." I believe OP has not done so, because of my understanding of WW2 in the Pacific. Ergo, it's not a "fully working" one in my judgment.



Title: Re: Found a fully working flat earth model?
Post by: jimster on April 29, 2022, 07:56:17 PM
I think we can actually agree on this.

If you bend the light however you need to, cancel the light where you need to (light comes off the sun only in certain directions), and ignore the problem of distance, then yes, the earth can be any shape.

At present, you have no explanation for why the same dome has Polaris everywhere, yet individual observers see it as directly north, while others see the entire dome as light blue with no stars, and yet others see an entirely different set of stars at the same time. You have no explanation for why longitude lines diverge when real life measurements reveals they converge.

RE geometry explains all this with conventional geometry, light travels straight in a vacuum, the sun radiates in all directions, and distance works consistently everywhere.

Are we all agreed on this?
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on April 29, 2022, 08:31:03 PM
Looks like neither of you read what OP has been writing. If you spend a few minutes actually reading, instead of knee-jerk reacting to yell "NO, FE BAD", you may just discover that the model is quite literally still the same model as the globe. Because it is still the globe model. OP spent a lot of time responding very clearly, and was very patient, and it's clear the folks that have had the most vocal opposition aren't even showing the slightest bit of respect by reading what they have written.
Title: Re: Found a fully working flat earth model?
Post by: existoid on April 29, 2022, 08:41:43 PM
Looks like neither of you read what OP has been writing. If you spend a few minutes actually reading, instead of knee-jerk reacting to yell "NO, FE BAD", you may just discover that the model is quite literally still the same model as the globe. Because it is still the globe model. OP spent a lot of time responding very clearly, and was very patient, and it's clear the folks that have had the most vocal opposition aren't even showing the slightest bit of respect by reading what they have written.

OP did not contend with the divergence of longitude, only that on either model the same place names on a map would be at the same lines of longitude, and that one can use the same type of calculation to derive those points regardless of the map projection used. My intrusion into the discussion adds a wrinkle not yet discussed in detail, I think, because the application of the plotting graphs in WW2 planes required the lines of longitude to not diverge.

My argument is a little more sophisticated than "NO, FE BAD" thank you very much.

Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on April 29, 2022, 09:04:04 PM
Looks like neither of you read what OP has been writing. If you spend a few minutes actually reading, instead of knee-jerk reacting to yell "NO, FE BAD", you may just discover that the model is quite literally still the same model as the globe. Because it is still the globe model. OP spent a lot of time responding very clearly, and was very patient, and it's clear the folks that have had the most vocal opposition aren't even showing the slightest bit of respect by reading what they have written.

OP did not contend with the divergence of longitude
Yes. (https://forum.tfes.org/index.php?topic=19093.msg257538#msg257538) Yes, he did. (https://forum.tfes.org/index.php?topic=19093.msg257575#msg257575)

Quote
My argument is a little more sophisticated than "NO, FE BAD" thank you very much.
It might have been, had you actually read what OP was proposing. This is still a globe, and those lines converge using the correct distance metric.
Title: Re: Found a fully working flat earth model?
Post by: existoid on April 29, 2022, 10:02:38 PM

OP did not contend with the divergence of longitude

Quote
Yes. (https://forum.tfes.org/index.php?topic=19093.msg257538#msg257538) Yes, he did. (https://forum.tfes.org/index.php?topic=19093.msg257575#msg257575)


Well, no he didn't, at least to my understanding.   I could be mistaken, but I'll explain why I don't think I am:

To cut to the chase, in discussing lines of lat/long troolon mentions that "The map has a different distance metric. Distance is just a formula, it's up to you to choose a meaningful one." And then in response to replies to that he writes, "For the AE map, the conventional distance formula for coordinates expressed as (lat, long) does work as lat/long are preserved by the projection."  And then later, "i'm trying to make a model of reality, i'm not trying to make a map.
- Coordinate transformations can turn any shape into any other shape
- physics works with coordinate trnasformations
-> physics can be made to work on any shape universe (have a look at http://troolon.com for pictures)
-> There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape."

And then further down, when SteelyBob presses on the question of distance, troolon writes (emphasis added):

"For an observer existing within the coordinate system, ie a person in austraiia, the world and distances appear as in reality.
For an observer outside of the coordinate system, you should measure distances with a flat-earth ruler. (which is curved and has non-equal distance markings)
Taking an orthogonal ruler, to a flat-earth coordinate system produces invalid results. Just like taking my bend ruler to your globe would completely invalidate it."

And voila, that goes directly to my initial point. WW2 planes used a plotting chart that required lines of longitude to not diverge, but to eventually converge at the south pole. If their plotting charts had matched the monopole map in reality, nearly all those pilots would have died in the ocean because they would not have known the absolute distances they needed to travel in miles to get back to their carrier or airbase.  Their plotting charts would have provided, in the words of troolon, "invalid results."

He very cleverly sidesteps the very different absolute distances between lines of longitude the further south they get from the equator if the world were monopole in reality, as compared to an oblate spheroid. He does this by discussing coordinates on map projections, and what he's written only makes sense (to me) under the assumption that the map projections preserve the coordinates as mapped to real world places, but not that the absolute distances as measured in miles would be the same between a flat monopole earth and a round oblate spheroid.

It's possible I'm mistaken about something troolon is saying, but I can't find any interpretation of what he's said in this entire thread that can be summarized as:

'Whether lines of longitude converge (as in RET) or diverge (as in FET), the absolute distances traveled between each line of longitude would be the same.'

That's nonsense, and I don't believe troolon made any claim to that effect!

Title: Re: Found a fully working flat earth model?
Post by: jimster on April 30, 2022, 01:05:46 AM
It is not still a globe model, literally. He has changed the shape of it. He is trying to say that because of his misunderstanding of math and physics, that any shape is the same as any other shape. A globe is a globe in any coordinate system. Changing the coordinate system does not make distances stretch or compress nor make the light bend. Coordinate systems describe geometry, a globe is a globe in cartessian, spherical, or any coordinate system and it has the same physical properties.

Math does no control, it describes. Physics is the process of looking at the world around us and figuring out a consistent explanation for how things work. Math is starting with a set of assumptions and building a logical system. Turns out that if you start with natural numbers and 3 space Euclidean math, you get a useful tool to help you do physics. But a globe is still a globe and has the physical properties of a globe (different from other shapes) no matter what equations you present. Changing coordinate systems does not change the shape of an object, or you have done it wrong.

If the transforms give equivalent results, you don't have to change the laws of physics to get observed reality, distance is constant, light rays don't have to bend and get cancelled to match observations.

Title: Re: Found a fully working flat earth model?
Post by: jimster on April 30, 2022, 01:08:38 AM
"-> There is no test to differentiate between the shapes. In reality we can only observe/measure the physical properties, not the shape."

If we can measure the physical properties, how can we not know the shape?
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on April 30, 2022, 12:31:58 PM
Great work, guys. You successfully demonstrated exactly what I said. Sort of, actually. Existoid read without comprehension, so that's at least more than was apparent before.

Every coordinate in 3D space is directly mapped from a single point on the globe to a single point on troolon's FE. It's a globe. Anyone standing on the surface of the Earth would perceive it to be a globe, which has converging lines of longitude at both poles.

But by all means, continue attacking this representation of the globe, expressed in a scary foreign coordinate system.
Title: Re: Found a fully working flat earth model?
Post by: existoid on April 30, 2022, 04:54:05 PM
Great work, guys. You successfully demonstrated exactly what I said. Sort of, actually. Existoid read without comprehension, so that's at least more than was apparent before.

Every coordinate in 3D space is directly mapped from a single point on the globe to a single point on troolon's FE. It's a globe. Anyone standing on the surface of the Earth would perceive it to be a globe, which has converging lines of longitude at both poles.

But by all means, continue attacking this representation of the globe, expressed in a scary foreign coordinate system.

Let me summarize troolon's point as I understand it, and as I understand you to be restating:

Troolon makes no claims as to which version of earth is a reality (globe or flat). However, whichever earth is "reality" - globe or flat - we can use a coordinate system to create a map projection to look like the other shape. In other words, if we assume the earth is a globe, we can use a 3D coordinate system to create an accurate map that looks like an monopole, it's just an AE projection of a spherical reality. Or, if we assume the earth is flat, we can use a 3D coordinate system to create a 3D model of a globe, but it's just a spherical projection of the flat reality. Of course you can use a 3D coordinate system to make any map projection, regardless of what the underlying reality is.

I'm not disputing that. You seem to think I am. I'm making a different argument:

If we assume that, in reality, the earth is a globe, and we map that globe using an accurate 3D coordinate system onto an AE projection (with the north pole at the center), and our AE projection is accurate to the globe reality, then regardless of unit of measure the absolute distances between each line of longitude at the circumference of the map is zero. That's because on an actual globe, such lines would converge at the south pole (which is represented as a big circle on this AE projection). The projection makes the lines of longitude appear to diverge, but if they diverged in reality then the distance between each could not possibly be zero. Two lines that are spaced apart can't have a distance of zero between them, can they?

Look at this image:
https://www.dropbox.com/s/ungp3c57f3ulcks/Monopole%20map%20with%20longitudes%20at%20the%20circumference%20highlighted.png?dl=0

Each red line would have a distance of zero between each other (no matter what units of measure you choose to use), if the world, in reality, were a globe, and this map was merely a projection of that real globe onto a flat surface.

By contrast, if we assume that, in reality, the earth is flat, with the north pole at the center, and we map that flat world using an accurate 3D coordinate system onto a spherical projection, with the outer circumference of the world condensed into a point at the "bottom" of the sphere, then regardless of unit of measure the absolute distances between each line of longitude which converge to a point at that "bottom" of the sphere  cannot be zero but a positive number!.

Remember, in this latter example we are assuming that the earth, in reality, is flat and so those red lines in my attached image are some measurable distance apart. Say, roughly 10,000 miles apart along the curve of the circumference. And that 10,000 miles of real ice between each is "projected" to a single point.

That's why I put an emphasis on this particular quote from troolon, which I repeat here without emphasis:

"Taking an orthogonal ruler, to a flat-earth coordinate system produces invalid results. Just like taking my bend ruler to your globe would completely invalidate it."

In other words, when you put actual, measurable numbers of distances between each line of longitude, the same coordinate system will produce invalid results depending on whether the real shape of the object measured is a flat monopole world or a globe. [EDIT: and thus my WW2 plotting chart evidence. The planes flying over the ocean had to have precisely measured distances between each stated coordinate - each line of longitude - on their charts, or else they would virtually always crash into the ocean and die. They couldn't just fly between unknown distances along each latitude until the next longitude without knowing the number of km they were traveling!].

I'll restate my final statement from my previous post a little differently given the fuller explanation above:

'The lines of longitude cannot both converge AND diverge in reality, because the absolute, measurable distance between converging lines and diverging lines cannot be the same. That's impossible. So, either longitude, in reality, converges (as in RET), but the AE projection using a 3D coordinate system merely appears like they are diverging when they're not. Or, lines of longitude, in reality, diverge further south (as in FET), but global representation merely make them appear like they are converging when they're not.

In reality, the same lines of longitude cannot both converge and diverge. They must be actually doing one or the other. Yes, the projection can make it look however you want.
Title: Re: Found a fully working flat earth model?
Post by: existoid on April 30, 2022, 05:29:01 PM

Every coordinate in 3D space is directly mapped from a single point on the globe to a single point on troolon's FE. It's a globe. Anyone standing on the surface of the Earth would perceive it to be a globe, which has converging lines of longitude at both poles.

I forgot to directly address is, although my above post certainly accounts for it.  But here's a more direct reply, taking into account my explanation above:

The "coordinate" that is exactly at the south pole in 3D space, when mapped to an FE map will become a circle. It will be the circumference of the world map, in fact. How can a point be a circle?

Or, if the reality were a flat earth, the many coordinates that form the circumference of the map would all be mapped onto the same single point that is the south pole in a spherical projection of the flat reality. How can a circle be a point?
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on April 30, 2022, 06:17:30 PM
In reality, the same lines of longitude cannot both converge and diverge.
I'm glad that you agree RET is an impossibility, but in this specific instance you happen to be wrong.
Title: Re: Found a fully working flat earth model?
Post by: existoid on May 01, 2022, 02:38:33 PM
In reality, the same lines of longitude cannot both converge and diverge.
I'm glad that you agree RET is an impossibility, but in this specific instance you happen to be wrong.

Ha, well put.

Let me clarify:

I should have written: "In reality, the same lines of longitude cannot both converge and diverge in the same direction moving  south of the equator."

Better?

[Edit: and in the paragraph above I could have been more specific as well, clarifying this topic sentence:
'The lines of longitude cannot both converge AND diverge in reality crossing through a series of given lines of latitude , because the absolute, measurable distance between converging lines and diverging lines cannot be the same.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on May 01, 2022, 03:14:38 PM
I should have written: "In reality, the same lines of longitude cannot both converge and diverge in the same direction moving  south of the equator."

Better?
But that's not what happens under RET (which I presuppose to be your definition of reality), and troolon's "model" is just a restatement of RET with no functional changes. His entire argument relies on taking RET piecemeal and throwing a layer of confusion on top of it.
Title: Re: Found a fully working flat earth model?
Post by: existoid on May 01, 2022, 03:38:33 PM
I should have written: "In reality, the same lines of longitude cannot both converge and diverge in the same direction moving  south of the equator."

Better?
But that's not what happens under RET (which I presuppose to be your definition of reality), and troolon's "model" is just a restatement of RET with no functional changes. His entire argument relies on taking RET piecemeal and throwing a layer of confusion on top of it.

Hmm.  Let me make sure I understand something:

You are saying that in the RET model, lines of longitude don’t converge as they go south and then meet at the South Pole? 

I understand they are also curving, in this model, but that’s along a different plane, and not really at issue for purposes of this discussion.

When I look at a 3D model of a globe and  at the lines of longitude south of the equator, they certainly narrow and narrow until finally converging at the South Pole.

What terminology do you suggest to be clearer in explaining how lines of longitude differ in the RET and FET (monopole) models than “converging” and “diverging” ?  (Referring to the lines as one moves south along them, to be clear).

Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on May 01, 2022, 07:03:49 PM
You are saying that in the RET model, lines of longitude don’t converge as they go south and then meet at the South Pole?
No, I'm saying they don't diverge, as long as we only consider what's happening south of the equator.

What terminology do you suggest to be clearer in explaining how lines of longitude differ in the RET and FET (monopole) models than “converging” and “diverging” ?  (Referring to the lines as one moves south along them, to be clear).
Let's be extremely clear here - troolon's "model" is not an FET model in any sense of the word, and it certainly is not a monopole model. It's RET with extra steps. It is nothing more, and it is nothing less. It is just RET, stated in a way that's confusing to some. If you think it is anything other than that, you are mistaken.
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on May 01, 2022, 11:53:38 PM
This is what I meant when I said you two hadn't actually read his posts. It's exactly the globe, expressed in a coordinate system that looks different without being different. The fact anyone is still suggesting the lines of longitude are diverging between the equator and either one of the poles in this thing he shared with us is clear evidence that the people making said suggestion haven't bothered paying any attention, yet they are still here to broadcast their ignorance and laziness for all to see while declaring some sort of weird victory. It's pigeons and chess.
Title: Re: Found a fully working flat earth model?
Post by: GreatScott on May 03, 2022, 03:56:20 PM
Quote
The map has a different distance metric. Distance is just a formula, it's up to you to choose a meaningful one.
Using the correct metric, the circumference of 80N and 80S is the same as the globe and so it's smaller than the equator

It sounds like the OP just made up his own metric.  That's not how it works.  You can't just choose whatever metric you want to use. Metric tensors transform according to specific rules and that's what determines the geometry.  When done correctly, according to the rules, the geometry of the manifold doesn't change.

Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on May 03, 2022, 04:11:36 PM
You can't just choose whatever metric you want to use.
Of course you can. In fact, you have no other option than to do so.

Metric tensors transform according to specific rules and that's what determines the geometry.
In Euclidean spaces, sure. This is emphatically not one. Considering you've missed that, I somehow doubt you know what you're talking about.
Title: Re: Found a fully working flat earth model?
Post by: GreatScott on May 03, 2022, 05:52:19 PM
You can't just choose whatever metric you want to use.
Of course you can. In fact, you have no other option than to do so.

Metric tensors transform according to specific rules and that's what determines the geometry.
In Euclidean spaces, sure. This is emphatically not one. Considering you've missed that, I somehow doubt you know what you're talking about.

Euclidean or non has nothing to do with it.  Metrics are tensors and tensors, by definition, are invariant under coordinate transformations.

Quote
A Tensor is an object that in invariant under a change of coordinate systems, with components that change according to a special set of mathematical formulae

If you transform the individual components correctly according to the transformation laws, the net result is the same metric that you started with.

If the OP has different metric for his globe model and his FE model, he didn't transform the components properly.

https://www.youtube.com/watch?v=j6DazQDbEhQ
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on May 03, 2022, 06:02:21 PM
Euclidean or non has nothing to do with it.
Right, I know everything I needed to know. Before you come back to argue against RET (which is what you're currently trying to do), get a grasp of geometry.

If the OP has different metric for his globe model and his FE model
He doesn't, and it's not a FE model. Please form an understanding of what's being discussed before you explain how proudly you disagree with it.
Title: Re: Found a fully working flat earth model?
Post by: GreatScott on May 03, 2022, 09:51:06 PM
Quote
Right, I know everything I needed to know. Before you come back to argue against RET (which is what you're currently trying to do), get a grasp of geometry.

 It doesn’t matter if you are talking about a Euclidean or non-Euclidean space, the metric tensor is invariant under  any  coordinate transformation.  If you are going from Euclidean to Euclidean, it is invariant.  If you are going from non-Euclidean to non-Eucldiean, it is invariant.  If you are going from Euclidean to non-Euclidean or the other way, it is invariant.

The whole point is that you can’t transform a Euclidean space to a non-Euclidean space accurately because they have different metrics.  The Euclidean metric is  the Pythagorean Theorem.  If you correctly transform to a non-Euclidean coordinate system, the metric remains the Pythagorean Theorem and distances and angles won’t make sense.  The PT doesn’t work in a non-Euclidean space.  You can’t randomly decide not to use the PT.
 
I guess you can decide that and just make up your own metric  and invent a bendy ruler to make it work.  The physics police won’t come and drag you away, but the results are meaningless.  It makes distances entirely subjective if you can arbitrarily decide to “measure differently”.  What makes his random metric the right one?  What makes his bendy ruler better than my super bendy ruler?  By the OP’s logic, I can change the defintion of a pound and claim I have lost weight. I still won’t fit into my skinny jeans though so it doesn’t mean anything.

Quote
He doesn't, and it's not a FE model. Please form an understanding of what's being discussed before you explain how proudly you disagree with it.

He specifically said they have different metrics.
Quote
The map has a different distance metric.
  And you may not consider it an FE model, but the OP does.  If you disagree with that then take it up with him.  I don’t care what the OP calls it. Or what you call it.  Either way, its not the right way to do coordinate transformation.
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on May 04, 2022, 12:47:16 AM
OP does not consider it a FE model. Try reading. If it doesn't work the first time, try again. He said more then once that it's still the globe. Other physicists reviewed it and said it's a globe (also, that's why it's "practically useless" - because using this framework is really impractical when you could use the actual globe). If you spend a few cycles pondering the implications of everything troolon described, it's obviously still a globe. But people get so hung up on fighting against a perceived enemy that it doesn't even matter what is actually being said.

Go ahead though. Tell us all how wrong it is, and how it can't possibly describe reality as we observe it, despite matching observations because it's, you know, the actual fucking globe and stars just illustrated in a way that looks scary and foreign and so, obviously, it must be bad.
Title: Re: Found a fully working flat earth model?
Post by: stack on May 04, 2022, 01:41:21 AM
I agree. It took me a while to wrap my head around it purely because the physical manifestation presented of a globe didn’t look like a globe. But at the end of the day, and this is a vast over-simplification, mathematically, creatively, and perhaps with newly devised measuring metrics instruments, you can make almost any shape represent a globe.

Even more simplified, just look at the dozens of globe projections that exist. None of them look like a sphere.
Title: Re: Found a fully working flat earth model?
Post by: GreatScott on May 04, 2022, 04:51:41 AM
Quote
Tell us all how wrong it is, and how it can't possibly describe reality as we observe it, despite matching observations

It only matches observations if you use a “bendy ruler”.  That’s like saying a movie perfectly reflects reality as long as you wear 3-d glasses.

Quote
OP does not consider it a FE model.

Then why did he title the thread  “Found a fully working flat earth model?” 

But like I said, I don’t care what he considers it or what you consider it.  I simply made the observation he didn’t transform the metric tensor correctly. It isn’t debatable that he did it wrong.  Tensors are invariant with coordinate transformations. Period. Full stop.

His two models, or  coordinate systems, if that makes you feel better, have different metrics. That means he didn’t transform correctly.  That’s tensor calculus 101….and basic differential geometry.  And we don’t even know if his “bendy ruler metric” is even valid. Valid metrics have defined characteristics.  You can’t just make up some random formula and call it a “metric”. 

I'll leave it to those who want to argue what the implications of those observations are.
Title: Re: Found a fully working flat earth model?
Post by: troolon on May 07, 2022, 10:00:06 AM
I believe there are still a few nuances in the way how i see this model versus how many posters view it.

Personally i don't think there has been enough emphasis on the fact that this model can be, and in fact was originally empirically derived.
2000 years ago, the Greeks made an incompleteness error: They assumed a globe and straight light and showed it was consistent with measurements in reality.
However, what they really measured was the relationship between light and earth shape. The earth shape was an assumption.

For my approach i've started over, but this time assuming a flat earth (later to be generalized to shape-agnostic)
I've started with the observation of horizon distance: ie that the formula R/cos(phi) - R   defines the relation between earth-shape and light-ray shape (basically height difference between a laser and a lake).
From this formula (and assuming a flat earth it's possible to create an explanation for day, night, seasons, .....)
In fact i believe it's possible to rederive all of physics this way without relying on the globe, at all.

At this point we have created an alternative model to the globe model and then there are 2 possibilities:
- we find a difference between both models and derive a test to see which one is correct
- both models are equivalent

In this case the latter can be proven. In history this has happened multiple times before. Think of the ptolemaic and tychonic models for planetary motion. It can be shown both are equivalent/approximations of the globe model)
In all posts so far, there has been a lot of emphasis on the equivalence with the globe model (as it's vastly easier to explain what's happening this way than having to rederive 2000 years of gnarly mathematics :)
Personally i do believe it is correct to call this a "flat earth model" (tough i admit there's a a bit of semantics involved)

Diving into this i've also discovered that the flat earth debate is a lot more nuanced than i initially thought. For example a picture of globe earth from space or a ship disappearing hull first are not actually proofs of intrinsic curvature for example and this is not something i would have guessed before this post.
And then there are of course the more philosophical questions about what it means to have 2 differently shaped models.

I also see this work as a potential bridge between RE and FE. When we see a ship disappearing hull first behind the horizon, and the globies and flatiies start warring whether it's the globe, the light or the aether that's curving, we now know they're really in agreement.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on May 07, 2022, 10:07:10 AM
I simply made the observation he didn’t transform the metric tensor correctly.
You don't know what that means. :)

Tensors are invariant with coordinate transformations.
Assuming a Euclidean space, which this emphatically is not.
Title: Re: Found a fully working flat earth model?
Post by: jimster on June 04, 2022, 12:17:20 AM
Troolon,

If you have oroved that the earth could be any shape, then the earth could be round. Interesting. The only thing I know all FEs to agree on is that the earth is not round. Could be north pole centric, could be south pole centric, could be bi-polar, but they all agree couldn't possibly be round. FEs will like you better if you prove it isn't round rather than that it could be round. To do that, I think you would have to prove that light rays in vacuum can't be straight and measurement is definitely broken. You will be the hero of FE.

I have a degree in math, my son is a math major, and I have reviewed the math behind your claims. Known mathematicians and discussed math all my life. Your claim might be restated as: The earth appears round in Euclidean 3-space but reality is secretly non-Euclidean (measurement is broken, as you say, and rays of light that seem to travel straight are actually bent, perhaps in different ways depending on the position of the observer). Once you assume measurement is broken and light curves in unknown ways, an infinite number of possibilities with no way to know which one.

You also did not incorporate the astronomical transform. When you apply your transform to earth, what about stars millions of light years away? Somehow they moved into a dome, but you have no math for that. Once you say the light bends and you don't know the equation, those stars could be anywhere, sun, moon, etc. The earth could be a cylinder a million miles long and an inch diameter. Meanwhile, Euclidean 3 space with RE gives us GPS, ICBM, airliner finds the airport, sextant north star latitude makes perfect sense, and the solution is singular. If light doesn't bend and measurement isn't broken, the earth is round.

Your ideas re coordinate transforms, basis, and "mathematically equivalent", and your physics claim that it follows a shape change through coordinate transformation are wrong. If you are right, you can go to the math and physics community and explain, for example, the power that coordinate transforms have to bend light. You will be famous, but you will need experiments and equations. You found a reasonable (to you, although not to a professor) of how the earth's shape is unknown. I imagine you will not share your discoveries with them, perhaps posting on TFES is the thing to do with it. To what end?

So the choice of what to do with your discoveries is up to you. You can post on TFES and get some agreement and some explanation of why you are wrong from RE. If I knew some part of science was wrong, I would go to scientists and explain in an attempt to set them straight. But I suppose if they did not agree, it would be because of conspiracy, or perhaps scientists and mathematicians are stupid. Conspiracy can really explain a lot, and you never have to have details or evidence, because its secret, so how could you know the details and the evidence is they all say the wrong RE stuff.

Still, just the idea that changing coordinates changes the shape of a geometric figure would be a starting point. Go to some mathematicians and show them how you changed a sphere into a disk by coordinate conversion. If they say "It doesn't work that way, you don't understand coordinate conversion, the figure remains the same size and shape." Perhaps all mathematicians are wrong, just the ones at UCLA, maybe you can explain, but this is their definition, so don't they get to decide?

You started with a globe where Australia's size matched real world measurements and Sigma Octantus made sense. Then you transformed the surface into a disk, where Australia was wrong size and Sigma Octanus made no sense. Then you claimed this as mathematically equivalent, and then that the physics is then equivalent.  Forget which is true, which is useful? If it pleasures you to think that under the RE appearance, non-Euclidean math means it is flat, well, I can't stop you. But the fact that you can map the points on a sphere to a disk if you don't care that the size is wrong and the light bends, but that doesn't prove the earth could be any shape. If anything, you have proved that it is round. In the infinite possibilties of shape agnostic earth, only one works with measurement and light waves traveling straight in a vacuum, Polaris and SIgma Octantus, etc, etc etc.

Does that mean anything? As I said, forget true, what is useful?
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on June 04, 2022, 10:59:17 PM
It's literally a RE but ok
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on June 06, 2022, 01:56:51 PM

Tensors are invariant with coordinate transformations.
Assuming a Euclidean space, which this emphatically is not.

This is incorrect. A true tensor is invariant under any coordinate transformation. A tensor is a multilinear map from vectors and dual vectors to real numbers and does not depend on a coordinate system to be defined. Its components in a give base may change when you change coordinates (and this happens whether in euclidian or non-euclidian spaces), but a tensor is not defined by its coordinates.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on June 06, 2022, 04:56:21 PM
A true tensor is invariant under any coordinate transformation.
I said nothing about coordinate transformations, merely the assumptions required for this statement to hold. Restating the statement louder is not gonna help.

Its components in a give base may change when you change coordinates (and this happens whether in euclidian or non-euclidian spaces), but a tensor is not defined by its coordinates.
Correct, but also delightfully irrelevant.
Title: Re: Found a fully working flat earth model?
Post by: jimster on June 06, 2022, 06:31:06 PM
The OP said that a sphere turns into a disk when you change coordinate systems. He said that the disk was mathematically equivalent to a sphere. A sphere is the set of points equidistant from a point. A disk is not that. There is so much wrong with the OP's ideas in basic high school geometry. Why even bring up tensors? Complexity is an FE tactic to make their idea possible by obscuring simple truth. The OP made numerous math errors ecxplained in the 14 pages above by myself and others.

Before any discussion of tensors, wouldn't it be wise to make sure the basic idea is not wrong?

Could we have a discussion of whether changing coordinate systems makes a sphere into a disk?

Does anyone else see the math errors? Crush me, show your math skills and prove to me that OP understands math correctly, coordinate systems, sphere, basis, and measurement. OP has changed all these things from math as taught in high school and college. Either all of math is wrong, or OP is wrong. OP wants to redefine math the way he needs to in order to prove the earth is flat, or at least could be flat.

The actual conclusion from the OP is that no one can know the shape of the earth because the light could bend in any direction and measurement could be broken. I am astounded that a site dedicated to the shape of the earth can end up talking about the math of tensors when the evidence of the shape of the earth is found in simple things. Only by ignoring those simple things can the earth be flat, so off to tensors and whatever.

You end up with 14 pages of people arguing esoteric math by people who mostly learned that math through a google search without a thorough background to understand correctly what those words mean. This is convenient for FEs so they can substitute their idea of what the math words mean so as to make FET possible. There are a few RE math knowledgeable people who try to convince them of the correct meaning, but that is like trying to teach a math course through a comment column to a person who does not want to learn the true math.

Troolon should go to a community college and enroll in a geometry class. Perhaps he could explain his ideas. Would be fun to see him explain how a coordinate conversion makes a sphere into a disk and yet is mathematically equivalent. Never fear, Troolon, you can save your belief system through conspiracy, just add mathemiticians to the NASA/RE conspiracy theory. Troolon correctly understands math, and the entire worldwide math community is made of a few conspirators and a bunch of stupid sheep.

Troolon right and they are all wrong? If they are right, then measurement works and light travels straight in a vacuum. If Troolon is right, measurement is broken and no one knows the shape and size of anything, not just the earth. How do you know about an object across the room when light bends and measurement doesn't work? In order to make it possible for the earth to be flat, you have to make it possible for anything to be any shape or size.

Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on June 06, 2022, 07:32:47 PM
A true tensor is invariant under any coordinate transformation.
I said nothing about coordinate transformations, merely the assumptions required for this statement to hold. Restating the statement louder is not gonna help.

Its components in a give base may change when you change coordinates (and this happens whether in euclidian or non-euclidian spaces), but a tensor is not defined by its coordinates.
Correct, but also delightfully irrelevant.

You didn't say, but you quoted specifically a part of someone else's post mentioning tensors are invariant under coordinate transformations, stating that this assumes Euclidian space. This is incorrect, tensors are invariant under coordinate transformations in any space they are defined, Euclidian or not. Perhaps my post wasn't clear enough
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on June 06, 2022, 07:43:18 PM
Complexity is an FE tactic to make their idea possible by obscuring simple truth.
This is your twice-pagely reminder that OP is a RE'er and he's making fun of all of you. There are no "FE tactics" in place here. OP is a RE'er and he is presenting RET, unaltered. It is extremely disingenuous of you to take a RE troll and describe his jokes as "FE tactics".

This is incorrect, tensors are invariant under coordinate transformations in any space they are defined, Euclidian or not. Perhaps my post wasn't clear enough
Your post was quite clear, and my response stands. If you think repeating yourself for the fourth time will help, it will not.

Perhaps your confusion lies in the fact that it is possible to come up with a non-Euclidean space in which your assertion holds. I could have been more precise, but I wasn't - it wasn't particularly necessary given that the exact space we're discussing has been specified, and given that the incorrect assumptions about this space were also made clear.
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on June 06, 2022, 09:58:54 PM
Complexity is an FE tactic to make their idea possible by obscuring simple truth.
This is your twice-pagely reminder that OP is a RE'er and he's making fun of all of you. There are no "FE tactics" in place here. OP is a RE'er and he is presenting RET, unaltered. It is extremely disingenuous of you to take a RE troll and describe his jokes as "FE tactics".

This is incorrect, tensors are invariant under coordinate transformations in any space they are defined, Euclidian or not. Perhaps my post wasn't clear enough
Your post was quite clear, and my response stands. If you think repeating yourself for the fourth time will help, it will not.

[/b]Perhaps your confusion lies in the fact that it is possible to come up with a non-Euclidean space in which your assertion holds.[/b] I could have been more precise, but I wasn't - it wasn't particularly necessary given that the exact space we're discussing has been specified, and given that the incorrect assumptions about this space were also made clear.

My assertion: Tensors are invariant under coordinate transformations.

Perhaps then you could help me clear my confusion please, because I'm having trouble coming up with an example that is the exact opposite of the bolded part in your quote.
Could you point me to an example of a tensor that IS NOT INVARIANT under coordinate transformation? Everything I'm reading says that this is not possible since the definition of a tensor does not rely on any coordinate basis, confirming my assertion.
Title: Re: Found a fully working flat earth model?
Post by: Pete Svarrior on June 06, 2022, 10:13:38 PM
Could you point me to an example of a tensor that IS NOT INVARIANT under coordinate transformation?
I don't know how many times I need to say that coordinate transforms are completely irrelevant here, but I'll try once more. If you keep rambling about this, I'll just ignore you.
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on June 06, 2022, 11:06:26 PM
Could you point me to an example of a tensor that IS NOT INVARIANT under coordinate transformation?
I don't know how many times I need to say that coordinate transforms are completely irrelevant here, but I'll try once more. If you keep rambling about this, I'll just ignore you.


Tensors are invariant with coordinate transformations.
Assuming a Euclidean space, which this emphatically is not.

But all of this started by you qualifying a claim that tensors are invariant under coordinate transformations, saying it assumes Euclidian space. Now coordinate transformations are irrelevant. Perhaps if you don't mean invariant under coordinate transformations, invariant with respect to what?
If you ignore it, ok no problem. We all will carry on with our lives
Title: Re: Found a fully working flat earth model?
Post by: Clyde Frog on June 07, 2022, 12:04:17 AM
For the sake of clarity: Do you think the OP is suggesting they are showing us a Euclidean space FE in this thread?
Title: Re: Found a fully working flat earth model?
Post by: spaceman spiff on June 08, 2022, 09:44:34 PM
For the sake of clarity: Do you think the OP is suggesting they are showing us a Euclidean space FE in this thread?

Don't know if this was intended to me or not, but OP's model is not Euclidian
Title: Re: Found a fully working flat earth model?
Post by: existoid on June 13, 2022, 11:22:02 PM

...words, words, words...

At this point we have created an alternative model to the globe model and then there are 2 possibilities:
- we find a difference between both models and derive a test to see which one is correct
- both models are equivalent

...words, words, words...


Troolon,

I'd love it if you addressed what I've written in this discussion.  I appreciate the replies to me by Clyde Frog and Pete, but my original comments and thoughts were all directed at you and what you've said, and I'm not entirely certain I've NOT simply been talking past Clyde Frog and Pete (and they've been talking past me in turn).

I have read your troolon.com page twice now, and I still can't see "where" it addresses the specific argument I made. No need for you to go back in this thread and re-read, I can succinctly repeat it here (and you'll see why I included your specific quote above).

Under your section headlined "Azimuthal transformation with the earth as origin" you have a sentence that reads:

"Around 6371km will be an azimuthal projection of earth (a flat earth)."

What does this sentence mean, exactly? I'm not sure I actually grasp what is happening in the transformation of the globe to a cylinder. The accepted radius of the earth is 6371km. Are you saying that the radius of the cylinder is that same distance of 6371 and therefore, the diameter of the "earth" (at height 0) is twice that distance, measuring 12,742km ?

If that's what you meant, then you haven't discovered an azimuthal FE model that works.

Look more carefully at the monopole FE model. If the earth indeed were shaped like a great disc, with Antarctica as a giant circle around the circumference, you'll see such a shape for earth requires a diameter (on that flat circle) of 25,484km. That means the radius for your cylinder concept would be 12,742km.  And not 6371km, as you state.

Am I making sense?  Or am I missing something?
Title: Re: Found a fully working flat earth model?
Post by: J-Man on June 27, 2022, 12:47:02 AM
My model is taught to my children and their friends. Please don't allow the satanic followers to sway obvious rational thinking.