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Messages - jimster

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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.

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

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 05, 2022, 01:10:41 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. 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.


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.

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

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

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

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

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

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 03, 2022, 03:41:35 AM »
The haversine gives the correct distance over the surface of a sphere between two points. When you transform this onto a 2d circle as in the FAQ map, you can have each point on the same lat/long as on a sphere. But those points are no longer on a sphere where the lat/long are both polar coordinates. Latitude is now along a straight line, you can't do trig with a straight line. Lat/long no longer has the meaning that makes haversine valid. All you are ever doing is getting the same answer as you would on a sphere. But you are no longer on a sphere, haversine means nothing when it is not segments of a curved surface. Of course you get the same answer, all you did was plug the same numbers into the same formula, but the formula does not apply to the product of the transformation. The distance must be calculated as the distance between the endpoints of the arc.

It is as if you took a flat map and determined a distance  by d = sqrt( (x2 - x1)squared + (y2 - y1)squared ), map it onto a sphere, label each point with the x,y position and then use the same formula with the same numbers, you get the same answer. But it isn't right, because the formula is only valid on a plane.

Meanwhile, in the real world, if you stretch latex over a sphere and paint the land masses of the earth on it, you can make a map with correct distances (haversone works). If you peel it off from the south pole up and flatten it, the southern hemisphere will have to stretch out. This is why Australia is half again as wide on the FAQ map. Now it is flat, though, and you need (angle. distance) and use 2d polar coordinate trig. Then you will get a number that matches the appearance/physical measurement of your transformed flat map. No stretchy ruler needed, it is a different distance, per Gauss.

Once again, haversine works on a sphere. When you flatten it, you have to use polar coordinate trig. Distance is not maintained through your transform. Latitude on FE map is not degrees on a sphere, it is just the relative distance from the center.  The haversine formula needs the angle between both latitude and longitude, but there is no such angle for latitude on a flat map.

Again, using haversine on a flat surface does not produce a right answer. . This is the error of your logic re "distance is preserved". Gauss says it is not, so do the faq maps that clearly show Australia bigger and Greenland smaller on the flat map. So does your model.

Where on your model is Sigma Octatus? Can you show startrails valid from every point? Can you show the entire dome visible to everyone, some with stars and some with daylight and sun? The entire dome filled with stars, yet completely different stars northern vs southern hemisphere? RET can, that's why the earth can only be round.

Distance is not preserved through your transform, RET accounts for all these things. Any other shape needs the light to bend due to "unknown forces with unknown equations". That's why we know the earth is round.

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.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: February 02, 2022, 09:09:11 PM »
Physics is looking at the world around us and developing a consistent explanation of what we see. Math is starting with some assumption and developing a logically consistent system. Physics is, math describes. The same physical object can be described in either cartessian or polar coordinates, but a sphere is still a sphere and a plane is still a plane. The coordinates are not bounded, 3 space to infinity is within both. The reeason locations on the globe are given as latitude and longitude, i.e. polar coordinates, is that the earth is round. If it were flat, we would use x/y coordinates, much easier.

If you transform a sphere into a plane or vice versa, points on the plane will be different distances, per Gauss. If you can make a flat map with constant scale, the earth is flat. If you can make a spherical map with accurate distance, direction, and scale, the earth is not flat, or Gauss' theorem is not true. There is no flexible measuring in geometry, the ruler is straight and constant. If you need to bend or stretch the ruler, you are proving Gauss' point.

When you flatten the globe into a disk as in the FAQ map, mathematically each latitude gets longer all the way to the south pole. Do car odometers in Australia measure distance differently from those in EU? Planes fly faster and have longer range? Do rulers stretch as they travel south?

Bending the light is, as the EA page in the FAQ says, "unknown forces with unknown equations". Making the ruler curved and the scale adjustable by location is fudge factor without any justification, what psychologists call "motivated reasoning". You get there by observing that assuming FE produces bad results, so you hypothesize fudge factor without proof in order to save your belief.

Your model shows how we could see sunset/sunrise and day/night on FE, assuming some directional phenomenon and the unexplained bending of light, coincidentally exactly the equations to transform RE into FE. But you're not done with explaining all phenomena we observe, we all see the dome, geometrically we all see all of it. Yet at the same moment, some see stars all over the dome, others see light blue all over the dome, they can be as close as perhaps 300 miles. Someone in the northern Hemisphere sees completely different stars than southern hemisphere.

Then there is Sigma Octatis, the southern (pretty close) pole star. At the summer (northern hemisphere) equinox, at midnight in South Africa it is just after sunset in western Australia and just before sunrise in South America. You can see Sigma Octatis directly south from all these places at that time. On FE disk map, Sigma Octatis is in directly opposite directions from South America and Australia.
So please show the dome appearance in your model. I will be interested to see: Where is Sigma Octatis and how do the light rays from it travel? How does the dome appear light blue for some, and for those who see it as dark, different stars. How do those stars appear to travel across the dome in different directions at the same time? Please show with your model.

Do you agree that flattening a sphere into a disk will geometrically distort the distances? Or is Gauss' theorem not true?

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: January 31, 2022, 03:05:28 AM »
My apologies, the graphics are small, but maybe you did take left/right into account.

You can't make Gauss Remarkable Theorem untrue by using polar coordinates. You still can't have 80 degrees north and south latitude both be smaller than the equator on a disk or a cylinder.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: January 31, 2022, 02:56:36 AM »
Can you make a graphic with the sun 93 million miles away and straight light path? Can you make a graphic of a globe map where all the distances and direction match observed? There are many many of these, I think you can. RE geometry works with straight light rays to explain day/night. Works without unexplained light bending.

The circumference of 80 degrees north or south is much less than at the equator. How does this project/transform/??? to a cylinder? The circumference at every latitude would be the same as the equator. To a disk? That's why Australia is always too large on the FAQ maps. This is possible on a sphere, how can 80 degrees north circumference be so much smaller than 80 degrees south on a disk FE? They should be equal. Per Gauss, you can't project, transform or ??? onto a cylinder or a disk and have both 80 degree circumferences be correct.

Your graphic of disk earth has Australia too big, just like the FAQ maps. Let's see your flat map with accurate size, distance, direction, and constant scale. You have math and graphics skills. Let's see it. Be the first!

Re day/night bendy light, do your equations account for the left/right bending? At noon on the equinox in Kampala, Uganda, the sun is rising in Rio De Janero. It appears to be directly east bearing 90 degrees, but on FE disk map, it looks like the bearing is actually about 45 degrees. And from Perth Australia at the same moment, looks like 270 degrees (directly west), but is actually 315 degrees.

Also interesting is that the light has no left right bending directly north and south. The amount of bend increases as you go south along the line of sunset until it gets to the far corner of the lighted area, where the discrepancy is maximum, then decreases until it is again 0 directly south.

So you need to add that to your model.

Interesting physical phenomenon, bends the opposite way on either side. As the sun moves around, this pattern moves with it. Apparently, the sun throws curveballs?

Now add the dome to your model. When it is sunset in Denver, in Salt Lake City it is still light blue over the entire dome. In St Louis, it is dark and there are stars over the entire sky, including the part of the dome beyond Salt Lake City. If someone in St Louis and someone in Salt Lake City look at the exact same spot over Denver, one sees light blue day sky, the other sees dark night with stars.

So let's see a model that accounts for left/right bending, and shows how someone can see either the entire dome of light blue at the same time as another person a few hundred miles east sees dark and stars, and has circumference of 80 degrees latitude both north and south much smaller than the equator on a flat disk, or cylinder, or anything other than a sphere.

You claim to have the ability to produce a FE map with accurate distance, size, direction, and constant scale. I await your post and will be astounded.

Flat Earth Theory / Re: Found a fully working flat earth model?
« on: January 30, 2022, 08:03:40 PM »
Unless you explain the curved light and the distortion of distances (per Gauss' Remarkable Theorem), can you really call it a "working model"?

That's why FE says nobody knows the distances over the oceans (per Tom Bishop), and the Electromagnetic Acceleration FAQ page cites "unknown forces with unknown equations" to explain the bending light. It is why all the maps in the FAQ have wrong distances.

So basically, the FE position is that light bends in exactly the way it needs to for FE but we don't know how or why and we can't know the info necessary to determine the the shape of the earth by distance measurements. Interestingly, this does not prove FE, the earth could be any shape, including round, and we can never know.

Oh, won't someone please figure out the exact way the light is bending and figure out how to measure the distances over the oceans? Perhaps we can never know?

Flat Earth Theory / Re: Where is Google Maps wrong?
« on: January 30, 2022, 07:13:41 PM »
Not gonna take the time to link to the video, sorry. It was a young couple in a 30-40 foot sailboat. The man was navigating with sextant only, did a noon sun shoot and calculated his position. Then he showed it to his wife, who checked gps on her cell phone. She then confirmed he was right to him - he wanted to navigate across the ocean without ever looking at gps. They edited in a chart showing the calculated and gps locations showing a few miles.

I make no claim to precision, but it doesn't take much precision to confirm that Australia is 2700 mi wide and US is 3000 mi wide and the FAQ map has Australia much wider than US. People on FE web sites frequently fixate on precision and miss entirely the point. It serves only to distract from the main point that proves the earth is round. For example here we could go off the rails to argue "round" vs "oblate spheriod".

Clearly people have been navigating the world successfully since 1700s. I flew on an airliner from Sydney Australia to LA. If you know how to do this, you know what the distances are, and hence the shape of the earth.

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