#### existoid

• 209
##### Re: Found a fully working flat earth model?
« Reply #260 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).

« Last Edit: May 01, 2022, 03:56:06 PM by existoid »

#### Pete Svarrior

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##### Re: Found a fully working flat earth model?
« Reply #261 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.
« Last Edit: May 01, 2022, 07:07:03 PM by Pete Svarrior »

<Parsifal> I like looking at Chinese Wikipedia with Noto installed
<Parsifal> I don't understand any of it but the symbols look nice

#### Clyde Frog

• 898
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##### Re: Found a fully working flat earth model?
« Reply #262 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.
« Last Edit: May 03, 2022, 07:27:33 PM by Clyde Frog »

#### GreatScott

• 24
##### Re: Found a fully working flat earth model?
« Reply #263 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.

#### Pete Svarrior

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##### Re: Found a fully working flat earth model?
« Reply #264 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.
« Last Edit: May 03, 2022, 04:24:35 PM by Pete Svarrior »

<Parsifal> I like looking at Chinese Wikipedia with Noto installed
<Parsifal> I don't understand any of it but the symbols look nice

#### GreatScott

• 24
##### Re: Found a fully working flat earth model?
« Reply #265 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.

#### Pete Svarrior

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##### Re: Found a fully working flat earth model?
« Reply #266 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.
« Last Edit: May 03, 2022, 06:08:04 PM by Pete Svarrior »

<Parsifal> I like looking at Chinese Wikipedia with Noto installed
<Parsifal> I don't understand any of it but the symbols look nice

#### GreatScott

• 24
##### Re: Found a fully working flat earth model?
« Reply #267 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.

#### Clyde Frog

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##### Re: Found a fully working flat earth model?
« Reply #268 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.

#### stack

• 3005
##### Re: Found a fully working flat earth model?
« Reply #269 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.

#### GreatScott

• 24
##### Re: Found a fully working flat earth model?
« Reply #270 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.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #271 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.
« Last Edit: May 07, 2022, 10:15:29 AM by troolon »

#### Pete Svarrior

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##### Re: Found a fully working flat earth model?
« Reply #272 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.