Offline troolon

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Re: Found a fully working flat earth model?
« Reply #100 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 ;)
      \__/

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #101 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)
« Last Edit: February 04, 2022, 11:48:05 PM by troolon »

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Offline Tumeni

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Re: Found a fully working flat earth model?
« Reply #102 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.
=============================
Not Flat. Happy to prove this, if you ask me.
=============================

Nearly all flat earthers agree the earth is not a globe.

Nearly?

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #103 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.

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #104 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.
I am really curious about so many FE things, like how at sunset in Denver, people in St Louis see the dome as dark with stars, while people in Salt Lake City see the same dome as light blue. FE scientists don't know or won't tell me.

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #105 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.

I am really curious about so many FE things, like how at sunset in Denver, people in St Louis see the dome as dark with stars, while people in Salt Lake City see the same dome as light blue. FE scientists don't know or won't tell me.

Offline jimster

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Re: Found a fully working flat earth model?
« Reply #106 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.
I am really curious about so many FE things, like how at sunset in Denver, people in St Louis see the dome as dark with stars, while people in Salt Lake City see the same dome as light blue. FE scientists don't know or won't tell me.

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Offline Clyde Frog

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Re: Found a fully working flat earth model?
« Reply #107 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.

Offline Rog

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Re: Found a fully working flat earth model?
« Reply #108 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.
Quote
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://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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #109 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....

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Offline Tumeni

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Re: Found a fully working flat earth model?
« Reply #110 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.

« Last Edit: February 05, 2022, 10:08:28 AM by Tumeni »
=============================
Not Flat. Happy to prove this, if you ask me.
=============================

Nearly all flat earthers agree the earth is not a globe.

Nearly?

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #111 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

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #112 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...

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #113 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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #114 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.

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #115 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.

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Offline Tumeni

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Re: Found a fully working flat earth model?
« Reply #116 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....
=============================
Not Flat. Happy to prove this, if you ask me.
=============================

Nearly all flat earthers agree the earth is not a globe.

Nearly?

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Offline Clyde Frog

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Re: Found a fully working flat earth model?
« Reply #117 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.

Re: Found a fully working flat earth model?
« Reply #118 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.


 

Offline troolon

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Re: Found a fully working flat earth model?
« Reply #119 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:

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.