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Messages - jimster

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Flat Earth Theory / Re: New Star Map
« on: April 29, 2022, 06:58:41 PM »
Do I understand Metatron's explanation of Polaris appearing directly north of any observer on FE be summed up as: "I don't know anything about how other than vague speculation about projection and reflection, it just does."? Sort of like "unknown forces with unknown equations" in the FAQ re Electromagnetic Acceleration? We can add unknown reflective properties of the FE dome?

Meanwhile, is there any flaw in the RE explanation for where Polaris appears? It requires only the known tested properties of physics, the light doesn't have to bend, no dome with mysterious reflective properties? Does RE explain this with simple geometry? What is the reason to prefer an explanation that has no explanation rather than one that has explanation consistent with geometry and physics?


How does gravity work in your model?


What do all FEs agree on? I started studying FE in 2015, Rowbotham published in 1864.

Any prospect of convergence?

Wow. The stars orbit the sun. Can you explain the orbit of Polaris such that it appears directly north in northern hemisphere and not visible in southern hemisphere? Also Sigma Octantus for the southern hemisphere?

Can you show me a diagram of their orbits over a spinning disk FE?

If Antarctica is at the center, then Sigma Octantus must be directly over the center of FE at all times. That is not orbiting the sun. Where is Polaris?

Flat Earth Theory / Re: Wiki on aviation
« on: March 09, 2022, 07:26:23 PM »
Tom Bishop,

I want to tell you a bunch of confirmable details on who my friend is and how I know he told the truth, but I do not want to revesal private info so you can see who he is, I don't do that on the internet for obvious general reasons.

Canadian Automated Air Traffic System, known as CAATS, was developed by Hughes Aircraft of Canada, Systems Division (HCSD) (acquired by Raytheon Systems Canada Ltd. in 1997) for NAV CANADA, and MDA was subcontracted by HSCD in 1993. CAATS automates flight data processing between all major air traffic control facilities, managing flight traffic over 5.8 million square miles of Canadian airspace. The CAATS system provides the air traffic controller with all radar, flight data, weather and other information on integrated, high-performance workstations.

Even if I am making it up (I'm not), there are programming departments all around the world doing geolocation programs, navigation programs, etc. They are either using RE equations or FE equations. One of the two gives wrong answers. CAATS is one of them.

Metatron, if I understand correctly, you are standing at the center of FE holding one end of a chain with a "large steel ball" attched at the edge of FE and you expect that if FE is rotating (never heard a FE theory where FE rotates around the center - the sun and moon always move), the chain will not have tension because it is moving slowly? That chain is 8000 miles long and has its own effect on the physics, but let's assume it is weightless. I need to know the exact situation with the ball, its mass and is it lying on the surface, coefficient of friction.

I suggest you forget the chain. In general, a ball at the edge of a spinning disk has angular velocity. If the ball is heavy and the spinning is slow, it just sits there. For 8000 mi radius disk spinning at 1 rev/day, the speed at the equator is 1000 mph, much more at the edge, too lazy to calculate, but 1000 mph is enough. So we have a presumably heavy ball traveling at 1000+ mph at the edge of a disk. That ball will travel straight at any given instant, but the path is curved, so some force must be applied towards the center or it flies off the disk. At 1000+ mph with a heavy ball, gonna be a lot of force.

Another way to look at it is suppose you have a ball traveling at 1000 mph on the end of a chain you are holding. Will the chain just lay on the ground? No. If the chain is 100 ft long and 1 rev/day, not much force. If the ball is traveling at 1000+ mph, huge force.

In general, your thought experiment is meaningless until you put numbers to it. You can imagine the results any way you like, perhaps you like an interpretation that fits with your FE beliefs. Real calculations with observed numbers and Newton's laws are consistent with RE

Flat Earth Theory / Re: Wiki on aviation
« on: March 08, 2022, 11:15:41 PM »
Pilots say they are flying level when they are flying at constant altitude. The curvature is ignored for aspects that are local, and they talk as though the earth is flat for convenience. Curvature of the earth mainly needs to be considered for navigation, just as he read. Locally, FE = RE or close enough. At 1000 miles, big difference.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: March 08, 2022, 11:00:04 PM »
Turning the earth into a different shape and then ignoring the distortion of distance, bending the light rays however they need to be bent, is certainly possible, although no answer for where is Sigme Octantus? Incorporating that into the model produce a rotating bent light tube not consistent with any known physics except the need to explain FET.

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

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

Rube Goldberg.

Flat Earth Theory / Re: Wiki on aviation
« on: March 08, 2022, 10:40:15 PM »
I told my software engineer friend about FE. He told me he was the programmer who wrote the routing software for the Canadian Air Traffic Control System. The US uses airways, set routes like interstates. The Canadians wanted to route their planes straight from departure airport straight to the destination for each airplane (routing around collisions, of course). He wrote the program that figured out the distance and direction. He said he used the spherical geometry equations straight out if a textbook and they worked perfectly. Obviously, they did much qa, and the system has been in use for decades, many planes arrive where they intended to go daily. If the earth was flat, the equations would be wrong and the pilots would not find an airport where the software sent them. The longer the route, the more the difference.

You can do this yourself, actually. Do the spherical trig math to calculate the distance between two cities, for Tom Bishop, make that two cities on the same land mass, perhaps Beijing and Madrid. For some reason, Tom Bishop thinks gps doesn't work over water. The RE 3d trig answer will match google maps, airline schedule, time/speed/distance of airliner flight, lat/long per wikipedia, etc, and no evidence at all for the FE distance calculation, whatever that might be.

Find a discrepancy between any RE info sources or the math calculations and prove FE! I will be your disciple. And I will make a lot better video than the one you just posted. His production values, scripting, delivery, etc is just bad regardless of the truth of his content. FYI, this maybe shouldn't matter, but it does. People are more likely to believe well made videos, at least most people. Perhaps FEs perceive truth in amateurish, clearly non-expert videos.

The wiki says that one proof point of FE is that you can't see or feel movement, therefor the earth is stationary. I went on a cruise on a very big ship. In my cheap cabin (no window), there was no motion, sound, or visual indication of movement. Yet I am sure that it did move, as I went to sleep in one port and woke up in another, with no sense of movement. My question is whether it is possible for someone to be moving, yet because the movement is smooth, silent, and everything within my sight is moving the exact same way, is it possible for someone to be moving without being aware that they are moving?

Flat Earth Theory / What is going on with these observatories
« on: February 23, 2022, 10:16:19 PM »

He said there are 35 such observatories establishing the positions of satellites in 3 space. Are these part of the conspiracy? Or are they all just wrong? Bear in mind that this is part of the gps system, and it works.

Are all the web sites that say where the satellites are correct, if so they establish RE. If not, how does gps work so well? Giant conspiracy? They are all wrong, but somehow gps works in some supersecret way that is not RE?

Can we say anything about this other than the earth is flat, so the math must be wrong and/or conspiracy?

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 23, 2022, 10:06:33 PM »
UCLA Math/Computer Science 1975. Never used the math because I became a software engineer. I can barely remember the math, but I learned stuff like what a coordinate system is and can refresh my math memory with internet. I remember enough to see the wrong math you do when make it complicated in order to hide the wrong part. Arguing details is pointless, you are not here to learn, you are here to justify FE.

I talked about FE with a programmer friend. He said that he had written the nav software for the Canadian Air Traffic Control system and used 3d geometry, haversines, straight out of a math textbook. Worked perfectly, airplanes in Canada today arrive exactly where they are supposed to, FE math would be different and not work. He has a degree in math.

I would be astounded if there was a qualified mathematician or physicist that agreed with any of Troolon's math. I am not surprised that he cites secret experts. Confirmation without possibility of falsification.

I remember when my linear algebra prof started the lecture with "Today we are going to talk about how we know the earth is not flat, Gauss' Remarkable Theorem." The normal vectors on the surface of a flat disk are parallel, so curvature is zero. The normal vectors on the surface of of a sphere are not parallel, so the curvature is not zero. Find me a math prof who will disagree with this. There aren't any. There is a reason why you are sayoing this on FE site, say it on a math web site, or astronomy, or astrophysics.

What you did is simple, you peeled the surface of the sphere off from the south pole and stretched it out into a disk, like popping a balloon and stretching the balloon out into a disk. All the rest is just blather. Doing this stretches out the size of everything in the southern hemisphere. In your graphics, Australia is clearly bigger than North America. Measurement is measurement, and there are many ways to know the true sizes.

But distance is not your only problem. When you stretch it, the south pole becomes the circle around the edge. If you incorporate Sigma Octantus, as directly above south pole, it becomes a circle rather than a particular point. Even if this made sense, you have to explain why observers in the southern hemisphere see it as a small dot on the part of the circle and do not see the rest of the circle. A difficult to explain combination of bent light and directional light, much like the spotlight sun problem, but worse.

Except for one thing, Sigma Octantus is not directly above the south pole, it is a little over a degree off axis. So consider the Southern Cross. It is enough south that it is seen from everywhere in the southern hemisphere as being due south. It is much like the big and little dippers in northern hemisphere. Where is the Southern Cross? It appears everywhere as outward from the disk. Where is it really? That question has no sensible answer on FE.

According to your theory, since everything is equivalent, seems like I should be able to do the same math conversion using the south pole. This gives Sigma Octantus pretty close to observed re azimuth. still requires bent light ("unknown forces with unknown equations", per the FAQ) for altitude. But now Polaris, the north star, is everywhere around the edge. Also seems pretty arbitrary to start with the poles, how about your house in the center? Start with your house and peel the surface of the sphere starting with the point directly opposite on the spherical globe. The same transform can be done choosing any point at the center. If you choose a pole, one of the pole stars makes at least some sense. Start at the equator and neither makes any sense at all.

You made graphics to show a transform to several different shapes. Include the pole stars in these transforms and see how much sense it makes. Let's see graphics that include the Southern Cross and the Little Dipper, visible from the places and in the direction that matches observations. You can't do it. Distances, direction, and observed location of astronomical objects are all correct in RE graphics. They are not correct in any other shape.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 16, 2022, 01:58:37 AM »

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

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

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

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

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

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

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

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

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

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

Where is Sigma Octantus?

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

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

Like this:

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

Flat Earth Theory / Re: Found a fully working flat earth model?
« 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.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 11, 2022, 11:08:36 PM »

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?

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 11, 2022, 10:45:21 PM »

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?

Flat Earth Theory / Re: Found a fully working flat earth model?
« 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.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 07, 2022, 02:13:30 AM »
The OP is claiming that the distances and physics of his model work the same as RE. This is not true, a disk is not equivalent to sphere, the distances are different. What he claimed was changing coordinates included flattening a sphere into a disk. Light has to bend in ways that there is no evidence of. He claims the earth can be any shape, physics works the same, they are mathematically equivalent. This is not true.

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

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

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

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

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

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

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

Again, his conversion flattens the globe, while he claims he is doing a coordinate conversion. It is not the same geometric shape. A sphere is not a disk. You can map the surface of a sphere onto a disk, but that is not the same model.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 06, 2022, 09:23:51 PM »
From your web page:

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

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

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

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

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

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

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

Flat Earth Theory / Re: Found a fully working flat earth model?
« 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.

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