#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #80 on: February 03, 2022, 10:34:16 PM »
Quote
it is impossible to differentiate between the different shaped models. There exists no test, observation or measurement to find a difference.

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

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

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

So in summary,
- It's possible to make physics work on a flat earth representation of the universe.
- There is no test that can distinguish the flat earth representation from the globe representation
- The true shape of the planet can be flat, or it can be a globe (or a square, or ....) but we will never be able to prove it.
« Last Edit: February 03, 2022, 11:37:55 PM by troolon »

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #81 on: February 03, 2022, 11:35:19 PM »
Repeat with me, haversine only works on a 3d sphere, not a 2d disk. The correct formula produces the exact thing you see in the faq map and your model. No distance funny business needed.

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

I believe you may be presupposing an orthonormal axis in all your arguments. In an orthonormal axis you are right. But we're working with spherical coordinates even though they're not drawn the conventional way.
This entire discussion really boils down to: what axis must be used to represent the universe, and orthonormal is just one of infinitely many.
There are several practical reasons to prefer orthonormal, but a lot of people forget it's just one of infinitely many possibilities.

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #82 on: February 04, 2022, 02:41:32 AM »
I am not assuming an orthonormal axis, and am sorry I brought up Gauss. Here is the explanation with high school trigonometry.

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

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

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

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

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

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

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

The only thing you can know in absolute terms is that you exist to have the thought. But for day to day life, RE works, FE doesn't.
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.

#### Tom Bishop

• Zetetic Council Member
• 10735
• Flat Earth Believer
##### Re: Found a fully working flat earth model?
« Reply #83 on: February 04, 2022, 02:50:25 AM »
Occam's razor: the true explanation is the simplest. For FE to be true, light has to bend due to "unknown forces with unknown equations", gravity is all messed up, a million things. For RE, all you need is the known behavior of light and physics.

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

https://wiki.tfes.org/Moon_Tilt_Illusion

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

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

The general argument against this is that it's some kind of vague illusion with some kind of mechanism to cause this to occur and that we should dismiss discussion of it, despite that it is what would happen if light were not traveling in straight lines on large scales.
« Last Edit: February 04, 2022, 04:02:48 AM by Tom Bishop »

#### Gonzo

• 78
##### Re: Found a fully working flat earth model?
« Reply #84 on: February 04, 2022, 07:44:17 AM »
Sigh.

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

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

Perhaps you were doing it wrong?

#### AATW

• 6591
##### Re: Found a fully working flat earth model?
« Reply #85 on: February 04, 2022, 09:37:27 AM »
Sigh.

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

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

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

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

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

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

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

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

(Or you could use a string to satisfy yourself that the lines are indeed straight.
Tom: "Claiming incredulity is a pretty bad argument. Calling it "insane" or "ridiculous" is not a good argument at all."

TFES Wiki Occam's Razor page, by Tom: "What's the simplest explanation; that NASA has successfully designed and invented never before seen rocket technologies from scratch which can accelerate 100 tons of matter to an escape velocity of 7 miles per second"

#### Rog

• 69
##### Re: Found a fully working flat earth model?
« Reply #86 on: February 04, 2022, 01:57:07 PM »
Quote
The intrinsic curvature of my space is 1/R²
How did you calculate that?
Quote
If one of the principal curvatures is zero: κ1κ2 = 0, the Gaussian curvature is zero and the surface is said to have a parabolic point.
Quote
The Gaussian curvature is the product of the two principal curvatures Κ = κ1κ2
.
https://en.wikipedia.org/wiki/Gaussian_curvature#Gauss%E2%80%93Bonnet_theorem
Where are your principle curvatures?  It doesn’t look to me like there are any curvatures.

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

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

You are contradicting yourself.  You claim your model has intrinsic curvature, but, by definition, intrinsic curvature can be tested for and observed from “the inside”
« Last Edit: February 04, 2022, 02:32:45 PM by Rog »

#### Tumeni

• 3179
##### Re: Found a fully working flat earth model?
« Reply #87 on: February 04, 2022, 02:59:45 PM »
You just need to look out your window at the Moon and celestial bodies to know that the straight line nature of light on a celestial scale is incorrect.

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

=============================
Not Flat. Happy to prove this, if you ask me.
=============================

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

Nearly?

#### WTF_Seriously

• 1336
• Nobody Important
##### Re: Found a fully working flat earth model?
« Reply #88 on: February 04, 2022, 05:05:03 PM »
Haven't thoughtfully tried it yet, but I'd propose the easiest way to debunk Tom's moon tilt defense is to observe the moon in the dark either pre dawn or post sunset in order to know where the sun should be.  In the dark, I would think the background causing the illusion would be removed.  Not sure that's true or not.  Just a thought.

I drive to work pre dawn every day.  I did notice last week that the moon tilt was definitely directed at the sun. However, I need to get a clear morning when the moon is close to setting to really see what the moon tilt looks like across the entire sky.
I hope you understand we're maintaining a valuable resource here....

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #89 on: February 04, 2022, 05:20:44 PM »
Tom Bishop,

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

So where on the celestial dome is sigma octatus?

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

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

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

Where is sigma octatus?
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.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #90 on: February 04, 2022, 08:33:17 PM »
The AE map is a plane, not a sphere. If all you did was change coordinate system, you did not make a FE map. When you turn a sphere into a flat surface (assuming you don't want to map points on the sphere to the edge or bottom of FE, thus a cylinder), you are doing a projection of 3d onto 2d. Longitude lines stay the same (preserve distance), and latitude turns into radius, the angle turns into distance from north pole.
The surface of a sphere (shell) is a 2D object. Any point on it can be represented by 2 coords(lat, long). A disc is also 2D. No loss of dimensions or information. A sphere is 3D. In my model, the sphere is transformed into a cylinder, also 3D.
In formulas: on a sphere: coords are expressed as (lat, long, radius), On the cylinder they're expressed as (lat, long, height) (lat is rendered on a straight axis, height from bottom which equals radius).
The only difference between between the sphere and the cylinder is wether we draw latitude on a straight, or a radial axis. Math uses both representations interchangeably (eg: polar coords -> radial, sine graph -> straight).

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

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

BTW, here's an observer traveling the globe and showing what part of the sky is visible. It explains from where you can see polaris and the south star. (i already uploaded pics on how different people can see the south star from different places)

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #91 on: February 04, 2022, 09:00:41 PM »
Quote
The intrinsic curvature of my space is 1/R²
How did you calculate that?
1. By construction. If a sphere in spherical coords has curvature 1/R², then so has mine
2. by the bug walking a circle test and comparing circumference of the circle with 2πR
3. by the fact any distance between any 2 points on the flat earth with my distance metric, is the same as on the globe. No distance has changed, then how can the intrinsic curvature?
4. By flying a plane with the normals test
Not all coord transforms change curvature (eg: cartesian sphere: x²+y²+z²=R² versus celestial sphere: lat=[0-π], long=[0-2π], distance=R)
All i'm doing is representing latitude on a straight axis instead of a radial one. This doesn't change the intrinsic mathematical shape.
It's like claiming i broke physics because i drew my graph in red pen instead of a black one.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #92 on: February 04, 2022, 09:26:34 PM »
You just need to look out your window at the Moon and celestial bodies to know that the straight line nature of light on a celestial scale is incorrect.

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

Straight like this?

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #93 on: February 04, 2022, 09:31:27 PM »
RET explains this, diagrams available in science textbooks. Is there a corresponding diagram and explanation for FE?

Where is sigma octatus?
Yes, check my model. Anything physics can explain, my model can do because it's the same model.
Physics works regardless of shape.
FE proof = express everything in spherical coords; proof in RE; express result in celestial coord again.
(or you can derive all formulas by hand, but that will take a while longer)

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #94 on: February 04, 2022, 09:34:41 PM »
Where is sigma octatus?
In my model: express sigma octatis as (lat, long, distance) from the center of the earth
Draw lat/long as on a AE map and place that in a cylinder at height `distance`
ie long on a radial axis, lat on a straight axis, distance in the up direction.

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #95 on: February 04, 2022, 10:11:37 PM »
Troolon says: In my model: express sigma octatis as (lat, long, distance) from the center of the earth

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

* Polaris

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

* Sigma Octatus

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

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

AE map:

* Sigma Octatus viewed from Australia

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

* Sigma Octatus viewed from South America

Please explain where Sigma Octatus is.
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.

#### Gonzo

• 78
##### Re: Found a fully working flat earth model?
« Reply #96 on: February 04, 2022, 10:16:01 PM »
Troolon,

In your diagramme above, what would I experience if I was stood on the top of the tower block, looking at the person on the right? Would they appear to be a lot lower than they actually were in reality? (Let’s remove the boat for the moment)

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #97 on: February 04, 2022, 10:46:16 PM »
Further thought re the location of Sigma Octatus in the FE dome model. Seems like it would be on the dome, so ...

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

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

Looking forward to seeing how your model incorporates that. Where is Sigma Octatus? RE has consistent plausible location. FE does not.
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.

#### jimster

• 292
##### Re: Found a fully working flat earth model?
« Reply #98 on: February 04, 2022, 10:59:34 PM »
Also, in your model, you show the light bending such that it crashes into the ground where sunset/sunrise occurs. Do you have an explanation why another light ray coming off the sun at slightly higher angle would not hit the ground beyond the sunset/sunrise line? What limits the direction the rays come off the sun? If nothing does, why is the whole earth not illuminated 24 hrs/day?

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

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.

#### troolon

• 101
##### Re: Found a fully working flat earth model?
« Reply #99 on: February 04, 2022, 11:25:42 PM »
Please explain where Sigma Octatus is.
On the AE map, where is the southpole? It gets deformed into the circle around the edge of the disc.
In my map, the southstar also gets deformed into a circle. Once you have a latitude that's even slightly off of -90°, it will become a point again.
So in the image below, the southstar is the yellow circle, and you can see all 3 observers see the circle due south.
You asked me to drarw the most abhorrent case on the AE cylinder, you shouldn't be surprised it looks weird
(btw on the globe, the southpole has no unique coordinates: lat=-90, long=[0-2π] which explains this result)

Once again this FE model can explain anything the globe model can because... it is the globe model.