Offline troolon

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Re: Found a fully working flat earth model?
« Reply #140 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.

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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

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.

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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.

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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?
« Last Edit: February 09, 2022, 07:59:42 PM by troolon »

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #141 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.

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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.

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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)

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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.

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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.
« Last Edit: February 09, 2022, 09:16:42 PM by troolon »

Re: Found a fully working flat earth model?
« Reply #142 on: February 10, 2022, 05:40:17 PM »
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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.

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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.

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We now have 2 models: one with a flat earth, and one with a globe.
No, still only a globe represented differently than usual
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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.
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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.




Offline Rog

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Re: Found a fully working flat earth model?
« Reply #143 on: February 11, 2022, 04:40:52 AM »
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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.

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.

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.

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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.
« Last Edit: February 11, 2022, 03:28:28 PM by Rog »

Offline Rog

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Re: Found a fully working flat earth model?
« Reply #144 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.

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #145 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.



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

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #146 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?


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

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #147 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?
"Electromagnetic Acceleration" sounds so much more sciency than "bendy light".

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #148 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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #149 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.

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Offline stack

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Re: Found a fully working flat earth model?
« Reply #150 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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #151 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.

Re: Found a fully working flat earth model?
« Reply #152 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!

Re: Found a fully working flat earth model?
« Reply #153 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. 

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #154 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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #155 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

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #156 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.
"Electromagnetic Acceleration" sounds so much more sciency than "bendy light".

Offline Rog

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Re: Found a fully working flat earth model?
« Reply #157 on: February 13, 2022, 07:01:57 AM »
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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.

The lines are distorted because the PT can’t be used to measure distances on a sphere.

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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.

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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
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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.
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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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #158 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)

Re: Found a fully working flat earth model?
« Reply #159 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.