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walter.bislins.ch

Flat Earth Dome Model

Friday, April 6, 2018 - 22:00 | Author: wabis | Topics: FlatEarth, Interactive, Knowlegde | Comments(32)
Here is an interactive Flat Earth Model that can explain Sunrise, Sunset, Moonrise, Moonset, Moon Phases, Moon's apparent rotation, Sun's position on Equinox, Seasons, Solar and Lunar Eclipses, Star Trails, 24 hours Day/Night at the Northpole and Antarctica, Celestial Poles, why people south of the equator can see the same Stars rotate clockwise around a singe celestial pole at the same time at different continents. But the Model still has some unsolvable flaws.

The Model

Before you think I am a Flat Earther please read the Conclusion and the Purpose of the Model.

  • Intro
  • Eclipses
  • Equinox
  • DayNight
  • Poles
  • Stars
  • Reset
  • <<
  • Play
  • >>
  • 0

Description

Light-Bending: This Model shows how light rays from the Dome on the Flat Earth Model have to be bent to match the apparent positions of Sun, Moon and Stars and to produce the Tracks, Star Trails and Day-Night Terminator as observed in reality for each time and location on earth. Only by bending the light rays as shown by this Model it is possible that Sun, Moon and Stars can go apparently down below the horizon, while they are still above the Flat Earth.

Sun/Moon Tracks: During 24 hours the sky with the fixed Stars rotates about 1 degree more than 360 degrees. So in 365.25 days the star constellations at the same time are at the same place in the sky again. You can see this in the model by advancing DayOfYear step by step (place the cursor into the field and hit arrow Up or Down). The Dome grid will advance each day by about 1 degree.

If you advance the time by 24 hours steps, the Sun moves up and down between the Solstice lines during one year, causing the seasons. The Sun also moves left and right a bit so it traces a figure 8. This is caused by the tilt of the Globe earth axes against the Sun Ecliptic plane 23.44 degrees. You can see the tracks of Sun and Moon against the fixed star background (Dome Grid) by checking the options Sun Track and Moon Track. A desciption of the tracks is presented if you click the Eclipses button. These tracks correspond to what is observed in reality. The tracks are derived from the Heliocentric Model.

Retrograde Motion of Moon's Track: The Sun Track stays fixed on the Dome Grid. But the Moon Track slowly rotates retrograde against the Dome Grid and rotates one full rotation in 6798 days. This is due to the precession of the Moon Orbit caused by the distant Sun. Currently the Moon Ecliptic is such that the Track of the Moon extends the Track of the Sun North/South about 5 degrees. In about 3400 days from now the Track of the Moon lies inside the Track of the Sun about 5 degrees. This observation has no explanation in the Flat Earth Model but follows from the Heliocentric Model.

Eclipses: The intersection points of Sun and Moon Track are called Knots. There are 2 such Knots marked by a green dot. If Sun and Moon are exactly on opposite Knots, a Lunar Eclipse happens. If Sun and Moon are exactly on the same Knot, a Solar Eclipse happens (play Demo Eclipses from Step 6 on).

This Flat Earth Model can predict Solar and Lunar Eclipses. But it can not predict moons shadow on earth at Solar Eclipses or earths shadow on the Moon on Lunar Eclipses, because the required relative sizes and distances of Sun and Moon and the spherical shape of the earth are essential to compute the corresponding shadow paths. So the location on earth where the Solar Eclipses happen can not be derived from the Flat Earth Model.

Moon Phases and Orientation: The model shows the Moon Phases and the Orientation of the Moon with respect to the horizon at the location of the Observer. The apparent rotation of the Moon during the day is due to the fact, that the cameras up vector stays always perpendicular to the surface of the earth while following the path of the Moon. An Equatorial Mount for the camera or telescope does not produce such a rotation as it follows the Moon.

Equinox: This Model produces the correct apparent Sun positions at Equinox, so that the Sun raises at 6:00 AM due East and sets at 18:00 PM due West everywhere on earth.

Poles: This Model produces 24 hours day and night on the Northpole and Antarctica.

Heliocentric Model: In reality ovserved Tracks of Sun, Moon and Stars (Star-Trails), the Equinox and Solstice Knots and the Day-Night Terminator can not be derived from the Flat Earth Model itself. They have to be assumed without any cause or reason. This Model derives them using the Heliocentric Model. On the Heliocentric Model they are simply a consequence of Gravity in play and the angles between the orbital planes of Sun and Moon. You measure the positions of Sun and Moon on their Orbits at any time and all future and past positions can be calculated using Newtons universal law of gravity.

Shapes on the Dome: The shape of Sun, Moon and star constellations on the Dome have to be distorted exactly like shapes of the real Globe world are distorted when mapped onto the Flat Earth. So the Sun and Moon on the Dome sould be squeezed circles, bent along a latitude line of the Dome. This distorted shapes get corrected by the bending of light as shown in the model, so the observer sees perfect spheres for Sun and Moon and the correct star constellation shapes.

All features of this Model are derived from the Heliocentric Model to produce an almost working Flat Earth Dome Model.

Problems with the Model

Lightrays: There are no known physical laws that can bend light so much and in this exact fashion to produce the real observations as shown in this Model. There are an infinte number of possible lightray paths that could be choosen to model the bent light. There is no physical cause to choose a specific one. So this Model chooses Bezier curves to model the light rays. The Bezier control points can be adjusted with the magenta slider. The smaller RayParam ist, the stronger the curve, the bigger RayParam, the smoother the curve.

Impossible Light-Bending: Atmospheric effects can never bend light that much. Not even known materials for lenses can achieve this. Even if such bending could be achieved by the atmosphere or something else, I can not derive any constant density gradient to produce just the right bending to connect each location on the Dome to each location on the Flat Earth at the right time.

No South-Pole: The Southpole makes serius problems. There can be no Southpole Star, because on the Flat Earth Model this star has to be smeared around the whole border of the Flat Earth.

Light bending over Night-Shadow: To produce 24 hours Daylight on Antarctica the Sun rays have to be bent over a region of Night shadow to the observer.

Moon Phases and Orientation: The Moon Phases on the Flat Earth can not be explained by the interplay of Sun and Moon with respect to the Observer. The Moon has to have it's own light. The Moon Phases and its orientation are complicated and can not be modeled by a simple rule on the Flat Earth. The Model derives the Moon Phases and its Orientation, as they can be observed in reality, from the Heliocentric Model. The Orbits and Distances of Sun, Earth and Moon are essential to derive the Moon Phases.

Shadows of Eclipses: Although the Model could predict the date of Eclipses, the Shadows on Lunar and Solar Eclipses, and therefore the locations and times on earth, where they can be observed, can not be calculated with the Flat Earth Model, because to compute the shadows you need the full 3D positions and sizes of the objects. You can't compute them from flat projections. All projections onto planes or domes lose the third dimension.

Brightness and Heat from the Sun: How does the sun produce the right amount of light and heat for each observer? The inverse square law does not work here because the individual distances between Sun and Observers in the model does not reflect reality, so the brightness and heat would never match observations. The Sun would have to produce light and heat different for each location on earth without any physical explanation.

Purpose of the Model

The purpose of this model is to show to what extent a Flat Earth model with a Dome can produce observations that match reality. This can only be accomplished by strong light bending. The model also shows, that many observations of reality can not be modeled with a Flat Earth Dome model.

The basic idea of my model is: The Flat Earth is a projection of the 3D Globe onto a flat plane. What if we project 3D space with Sun, Moon and Stars (and Planets) onto a 2D Dome? That is what my model does.

Note that although the Dome itself may be 3D, it only represents a 2D surface. Of course applying the known physical laws of light propagation, on the Flat Earth we would see a completely different imgage of reality than we can observe. Sun, Moon and Stars on the Dome never go physically below the horizon. So you have to invent things like the Flat Earth Perspective, which does not work as needed either. But if you assume light bending as shown in my model, it can really produce the images that we observe to a certain extent, if you don't look too close into it. Exceptions are Solar and Lunar eclipses for example. Although you can predict the dates of this events, like the ancient astronomers could by observing the sky, but you can not predict the locations on earth, where this events happen. They can only be seen on certain locations and times, which we can only predict using the Heliocentric model.

Some other points I wanted to show with my model are:

Sun and Moon trace specific paths on the celestial sphere. This paths have no cause on the Flat Earth model. The only explanation is, a creator has created it that way. But in the Heliocentric model all paths follow automatically from the law of universal gravity. You only have to measure the current positions, velocities, sizes and distances of Sun, Planets and Moons and you can calculate all past and future locations and how they appear from the earth or any other place by applying the law of gravity alone. You can predict the exact locations and times where solar eclipses can be seen (from the shadow that the Moon traces on earth).

The Moon phases and its apparent orientation for any observer on any place on earth, as shown in my model, can not be explained by the Flat Earth model. They have no relations to the flat earth sun too, so the moon has to have "its own light". My model can predict this observations for any place on earth by using the Heliocentric model.

The Day/Night Terminator on the Flat Earth that matches reality has a very peculiar shape that changes over the course of a year. The shape depends somehow on the location of the sun. This shape can only be explained with the Heliocentric model by projecting the Terminator from the Globe Earth with a tilted axis onto the Flat Earth. The shown light bending in my model would produce this terminator line correctly. But this light bending is not a thing that arises naturally, but is computed explicitly in a way to produce the real observations.

To have a Dome means, all heavenly bodies are located on or near the Dome. The real solar system is a 3D space with big objects very far away from each other orbiting around each other. The Dome is a 2D projection of this 3D space, very similar to the 2D projection of the Globe Earth onto the Flat Earth plane. This produces inevitable distortions you have to correct somehow (by bending light). By loosing the third dimension you loose a lot of information to produce and predict real observations. So Flat Earther have to "make things up" to explain observations that follow automatically from the real 3D universe we live in.

Conclusion

Some observations like the positions of Sun, Moon and Star Constellations as well as Sun/Moon-Rise and -Set can be explained by a Flat Earth Model if we allow strong light bending in a specific way. Even the date of Eclipses can be predicted from this model.

But observations as the southern celectial pole, Moon Phases and its Orientation and apparent Rotation as well as the track of the Shadow of the Moon on solar Eclipses can not be computed from the Flat Earth Model, because you need the correct sizes and orbits of Sun, Moon and Earth to compute this. The essential third dimension is lost if you assume a Dome over the Flat Earth, where Sun and Moon are close and small.

There is no explanation or scientific model that can physically explain why and how light is bent as needed by this Flat Earth Model.

Last but not least, the Flat Earth does not represent the real shapes and sizes of the continents. This can never be accomplished. A 3D sphere, as the Earth really is, can never have a similar surface as its flat projection. This is geometrically impossible. That's the reason why a sphere looks different than a plane in the first place, because their surfaces have different curvatures. They can never match globally. You can only make small flat map projections of the Globe surface that get not distorted too much to be usefull for finding local places. But you can never accurately measure real distances from any flat map projection. Global navigation has to, and always did, use spherical coordinate systems, like the current WGS84 model, used by GPS, aricraft navigation systems and google earth.

Equinox, Solstice, Azimuth, Elevation

The Model implements a perfect circular Orbit of the Earth around the Sun and a perfect circular Orbit of the Moon around the Globe Earth. This results in a slight divergence of the dates of Equinox and Solstice from reality of a few days.

The Model chooses to match:

  • Spring Equinox at 20. March 2017 12:00 UT
  • Solar Eclipse at 21. August 2017 18:00 UT

Azimuth and Elevation of Sun and Moon are also slightly inaccurate due to the use of circular orbits instead of elliptical orbits. This affects also the Moon Phases.

Computing Day-Night Terminator

The Day-Night Terminator is derived from the Heliocentric Model to match reality as follows:

  1. A circle perpendicular to the axis Earth-Sun in the Sun coordinate system is computed depending on the Sun's position at a certain date and time relative to the intersection knot of the equatorial plane of the earth and the ecliptic plane of the Sun.
  2. This circle is then transformed to the Globe Earth coordinate system.
  3. Each point of the circle is then transformed to the corresponding Flat Earth coordinate system, which produces the shadows as shown in this Model.
  4. At Equinox the transformation to the Flat Earth coordinate system fails at the south pole, because on Flat Earth the south pole is everywhere on the border of the disc. The corresponding points of the Terminator line had to be generated explicitly and have no real cause.

The special shape of the Night-Shadow produced by the mapping of the Globe Night-Shadow onto the Flat Earth results automatically, if the light rays are bent as shown in this model.

Computing Moon Phases

The Moon Phases and their orientations with respect to the Horizon at the Observer can only be computed from the Heliocentric Model as follows:

  1. The relative positions of Sun and Moon from the Heliocentric Model with respect to the Geocentric coordinate system determines the Phase of the Moon and its axes as observed from Earth. The distances of Sun and Moon from the Earth have to be correct to produce the correct phases.
  2. The Coordinate System of the Observer's Camera is computed in such a way, that the Camera points to the Moon and the Up direction of the Camera is perpendicular to the Globe Surface at the Observer.
  3. The Axes of the Moon Phase is transformed to the local Coordinate System of the Observer.
  4. The Orientation of the Moon Phase is computed from the Axes in the local Coordinate System of the Observer by transforming it to the Camera Coordinate System.

Used distances:

  • Earth-Sun: 149,600,000 km
  • Earth-Moon: 384,000 km

URL Parameters

The App can be called with some URL parameters to set a specific App State:

You can use the Save/Restore Panel with the Button Get App URL to get an URL from the current state of the App. Copy this URL into any Web-Form (e.g. YouTube comment) and by clicking the link this page is opened with the saved state from the URL.

  • To start a certain Demo with the URL, append
    &demo=DemoName to the page URL, where DemoName is the Name of some of the Demo Buttons above the App Dispalay.
  • To set a demo to a specific position, append
    &demo=DemoName&pos=number.
  • To run a demo from a specific position, append
    &demo=DemoName&play=number.
  • To speed up or slow down all animtion, append
    &speed=SpeedFactor.

Some parameters may be combined in one URL, eg. &demo=Intro&play=3&speed=3

The App changes the URL in the adressbar of the browser to reflect the current state of the App. You can copy the URL and post it in a comment. With the Back and Forward browser buttons you can jump from state to state.

Comments

1Dave 4/7/2018 | 17:25

Please do this with a geocentric model as the flat earther theory is inside the dome and if any flat earther tries to say that it is possible that the sun is outside the dome, they then flaw every attempt to prove its flat

2wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 4/8/2018 | 16:33

Sun inside or outside the Dome (Wabis to Dave)

It is not relevant for this model whether the sun is inside or outside of the dome. The dome shown in this model is simply a visual grid to show the locations of the sun and moon. You can imagine the real dome to be farther out if you wish. That does not change anything.

By the way: you can modify the height and size of the dome with the blue sliders.

3Greg McCormick 4/10/2018 | 10:13

Thoughts about Refraction (Greg)

Thank you for this splendid piece of work. I hope you will use it as the basis for some videos.

Your model should also force the correct elevation of Polaris by latitude. I tried to derive a vertical profile of refractive index that would mimic the observed relationship. I gave it a boundary condition that the refractive index should match that of air up to about 20000 metres. By having a varying refractive index above this height (in the region of 1.2) it is possible to get a match plus or minus 5 degrees. So far I have not been able to do better, and I suspect it may not be possible to do better with these boundary conditions but I have not yet proved this.

There is an amusing consequence of the bending. If this is due to refraction, all flat earth estimates of sun height must be wrong except those made at the equator, and all flat earth estimates of Polaris's height must be wrong except those made at the North pole. And those two "heights" of course exactly match the known radius of the earth.

4Jason 4/25/2018 | 19:46

The Earth has already been irrefutably proven to be flat and stationary. The smart phone in your pocket provides proof by itself. So any phenomena must be put in context of a flat, stationary Earth. Your theory of bending light is definitely an intriguing approach to take, and your animating model is much appreciated. Great start for a flat Earth model in motion using your theory!

Requests (Jason)

Provide a setting to show how the light paths from one star would be bent in many directions at the same time, along a single Latitude line in the night time areas of the Earth, so we can see the three dimensional shape of all the curved lines together

Provide a setting to show how the light paths from the Sun would be bent in many directions at the same time, forming the daylight areas of the Earth, so we can see the three dimensional shape of all the curved lines together.

The 'celestial sphere' seems to be much too big. Allow the user to change the size and proportion of the 'celestial sphere'.

Some Considerations (Jason)

Magnetic field lines of force are all curved. Dielectric field lines of force are all curved. Light can be bent dramatically by electromagnetic fields. The Earth has field lines of force, including the lay lines.

The 'firmament is not a dome, but instead a volumetric field, the bottom of which is about 70 miles altitude above us.

The consideration of any atmospheric refraction of any light from above would be a 'last second' distortion that might be ignored for this particular model.

In your descriptions, consider leaving out any reference to the heliocentric model because it's irrelevant for this experiment.

It's possible the Sun is not a bright ball but instead an electromagnetic projection created by crossing field lines of force, so the Sun appears in a different, predictable spot in the sky depending on your location.

Thanks for taking the time to put this together. Really cool.

5wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 4/25/2018 | 22:09

Answers to Jasons Considerations (wabis)

Jason
The Earth has already been irrefutably proven to be flat and stationary.

On the contrary. There is not a single observation that is not also a prediction of the globe model but uncountable observations that contradict the flat earth model. Not to speak of the technical achievements we have like airplanes navigating using GPS satellites and inertial reference units that depend on gravity, time and the rotation of the earth to function. Only to name some.

Jason
The 'celestial sphere' seems to be much too big. Allow the user to change the size and proportion of the 'celestial sphere'.

The size of the celectial spheres has no influence at all. It is a virtual image as it appears to the observer. I don't want to add too much features. But thanks for the suggestions. Maybe later...

Jason
Magnetic field lines of force are all curved. Dielectric field lines of force are all curved. Light can be bent dramatically by electromagnetic fields.

This is a wrong conclusion without any experimental support. In fact mainstream physics (Maxwells theory of EM) and all experiments tell us, that light is not bent by electric and magnetic fields. Try to bend a light ray with a strong magnet. In old TV tubes there are extremely strong electric fields, but light passes this fields unaffected. Light does not interact with itself either (except in very special circumstances we never experience on earth).

Before light gets influenced by any electric or magentic field, they have to be so strong, that their energies would destroy everthing that comes close to it. Atoms get smashed befor light is affected.

Jason
The 'firmament is not a dome, but instead a volumetric field, the bottom of which is about 70 miles altitude above us.

Evidence? Did you send ballons up there to check this hypothesis? We send rockets into space almost daily and never found a dome of any kind.

Jason
In your descriptions, consider leaving out any reference to the heliocentric model because it's irrelevant for this experiment.

It would suit you that way. Flat Earthers tend to deny things that do not fit their believe or imagination of reality.

The Heliocentric model is very relevant! One of the main purposes of this model is to show, that most celestial observations can only be explained with the Heliocentric model. Some aobservations are even impossible on a FE model, like the fact that solar eclipses are local phenomena, because the shadow of the moon produced by the distant sun crosses only a small parts of the earth. The Moon phases as shown in my model, matching the observations to a precision of some minutes, are computed from the Heliocentric model. I have no way to compute it from the FE model.

Jason
Thanks for taking the time to put this together. Really cool.

Thank's for that!

6Ride79 5/1/2018 | 22:14

Wondering if it’s possible that the sun is the focal point of a light being magnified through said dome.

7Jason 5/3/2018 | 17:55

MEMS (Jason)

The MEMS gyroscope and MEMS accelerometer ignore all forces, including gravity. A lot of people who believe in the space balls model claim MEMS technology measures gravity, but when you ask them how that is possible, they change the subject every time. Get the sensor kinetics app, put your phone on a level table, drive 110km, put your phone on another level table, and see how much curvature you drove over. 0°. The Earth has no curvature. Everything must be put into the context of a flat, stationary Earth.

Ionosphere (Jason)

The 'ionosphere' is an invisible barrier about 70 miles altitude that nothing physical can get through. More than likely it is an electromagnetic barrier. Everything in our existence is electromagnetic.

The Sun and Moon are above that barrier, so we will never get to them. All we can do is surmise based on observations. If you can make your model a little more interactive per my previous suggestions, then we can begin to see the overall shape of the bending light and compare it to invisible electromagnetic field lines of force that we can see using simple instruments.

I'm not afraid of the results, and you shouldn't be either. I know you are the one putting time into this, so you are the gatekeeper when it comes to the information you are willing to explore.

REQUESTS

1) Provide a setting to show how the light paths from one star would be bent in many directions at the same time, along a single Latitude line in the night time areas of the Earth, so we can see the three dimensional shape of all the curved lines together

2) Provide a setting to show how the light paths from the Sun would be bent in many directions at the same time, forming the daylight areas of the Earth, so we can see the three dimensional shape of all the curved lines together.

Thanks again.

8wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 5/3/2018 | 19:57

How MEMS work (wabis)

Jason
The MEMS gyroscope and MEMS accelerometer ignore all forces, including gravity.

Certainly not. How can anything ignore forces at all? How do you come to such claims? What is the purpose of a force measuring device that ignores all forces? Really?

Jason
A lot of people who believe in the space balls model claim MEMS technology measures gravity, but when you ask them how that is possible, they change the subject every time.

So then I try to explain it to you: Gravity is an acceleration, beacuse all objects in free fall are accelerated by the same rate of 9.81 m/s2 towards the center of the earth. How does a MEMS measure acceleration? It contains a testmass suspenden on springs. If you accelerate the MEMS case, the testmass want's to retain its state of inertia. So the accelerating MEMS has to exert a force to the mass m to accelerate it with it. This force can be measured by the MEMS somehow. From this force F the applied acceleration a can be calculated by a = F / m. So if you accelerate a MEMS horizontally, the force needed to push the mass to gain the same speed as the MEMS is proportional to the acceleration. By measuring this force you get the acceleration.

If the MEMS is laying on a table, its gravitational acceleration is prevented by the upward force the table exerts to the MEMS. But the testmass feels the gravitational acceleration too. It must be prevented from beeing pulled down by gravity by a force the MEMS is applying to the testmass. This force F = m · g is proportional to the gravitational acceleration g. So from the perspective of the MEMS, it looks like the MEMS is constantly accelerated upwards by the force the table exerts on the MEMS. The deeper underlying principle is the Equivalence principle of General Relativity.

So a MEMS does in fact measure the gravitational acceleration directly. But it shows it as an upward acceleration, because from the perspective of the MEMS it looks like the table is accelerating the MEMS upwards to compensate the downward gravitational acceleration. The net effect of the two accelerations is zero, so the MEMS stays where it is, but feels an upward acceleration due to gravity. Look at a full motion cockpit simulator. They imitate forward accelerations by tilting the simulator backward, so that part of the gravitational acceleration pulls you backwards and you have the illusion you get accelerated forward by the seat you are sitting in.

Jason
Get the sensor kinetics app, put your phone on a level table, drive 110km, put your phone on another level table, and see how much curvature you drove over. 0°.

Now that you know how a MEMS measures gravity, it sould be easy for you to recognise that you can put your phone anywhere on the globe onto a table and it would everwhere measure an acceleration exactly upward in the opposite direction of gravity, which always acts to the center of the Earth. So the angle between the local horizon and gravity measured by your phone is everywhere on the globe Earth the same 90°.

One might think, that the gyros in MEMS could measure the rotation of the earth or the change in your oriantation with respect to space when you travel. But the MEMS gyros are not mechanically rotating discs which are rigid in space. They can't measure absolut orientation. They measure angular speed. In pinciple you should be able to measure the angular rotation rate of the earth with MEMS gyros. The angular rotation rate of the earth is only 0.25 degrees per minute or 0.004167 degrees er second. The typical resolution of a MEMS is about 0.0038 degrees per second but the signal noise is even greater, about 0.014 degrees per second. So this sensors are not able to measure earth's rotation or absolute orientations even approximately. They are designed to measure high rotation rates of phones, which are in the range of at least some degrees per second. They can be initialized to measure a range of up to 2000 degrees per second. The smallest range is up to 125 degrees per second. Compare this with the needed at least 0.004 degrees per second. No chance! https://www.bosch-sensortec.com/bst/products/all_products/bmg160

Conclusion: MEMS in your phone can't be used to detect earth's motion in any way. The are not stable and accurate enough.

Ionosphere is not a barrier (wabis)

Jason
The 'ionosphere' is an invisible barrier about 70 miles altitude that nothing physical can get through. More than likely it is an electromagnetic barrier. Everything in our existence is electromagnetic.

The Ionosphere is not a barrier except for some radio waves. The Ionosphere is a region of very, very diluted ionized gas. This is nothing that can stop anything solid. You can move through atmosphere thousands of times denser without being stopped.

Where did you get all this misinformations from? Flat Earth videos I guess. Tip: use https://en.wikipedia.org/ as a starting point for your research instead.

Jason
The Sun and Moon are above that barrier, so we will never get to them.

We were already on the moon and have currently probes on it and in its orbit. And this year starts a satellite that will fly-by near the sun like no satellite ever before, inside the orbit of mercury. The Parker Solar Probe: Our First Mission to Touch the Sun! from Deep Astronomy.

Json
All we can do is surmise based on observations.

No, we can not only observe from the earth, we know how physics work by observations and experiments. We figured out that physics works the same everywhere and we made use of this knowledge. So we could actively send probes first on balloons to the hight atmosphere then with rockets higher into space. So we know what outer space is, we don't have to guess any more. We know how to land on other moons and planets and did it many times already. The deniers do simply not understand how things looks like in space conditons. They look not the same as on earth! And from the lack of their understanding they conclude it can't be real, must be fake. I feel so sorry for this people because they have to deny the infinte beauty of outer space and call all liers that know about that.

Jason
If you can make your model a little more interactive per my previous suggestions, then we can begin to see the overall shape of the bending light

Jason, I already implemented your suggestions, see options RayTarget = FlatEarth and RaySource. Check out all Demos (click the buttons above the App), the new features are incorporated into the Demos.

No light bending by EM (wabis)

Jason
...and compare it to invisible electromagnetic field lines of force that we can see using simple instruments.

I have to dissapoint you. Light is not bent by electric or magnetic fields. Have you ever seen a lightray be influenced by an electric or magnetic field? The only things that can bend light are density changes or gradients in gas (very little), liquids or solids, and by very strong gravity (spacetime bending).

There are no known physical effects that can bend light in such peculiar ways, needed to make a flat earth model plausible at least a little bit.

I recommend you, study real physics - in courses, not Youtube Videos. It's the hard way but the only one that works. Get an understanding of how things work. We figured it out since centuries. The knowledge is all there. Recognize the real shape of the earth and the universe. Then you suddenly are able to go to the moon if you want. It is possible, real, and done already many times. At least you could see the lies the flat earth gurus tell you, making money from people who don't know anything and are gullible.

9Ivo 5/6/2018 | 10:10

Hm... I am trying to see how equinox is working, but can't understand it properly. Does sun rays have different bending for any observer, for example due east from my position (45°N) and different bending, for example due south, and what happens during the day on equinox? I tried to adjust position of the observer on your model, but is not working, or I can't understand how to use this model of yours.

I really appreciate your effort to show mathematically imposibility of flat earth model, but I am curious about equinox case because if we calculate sun positions it comes out that every observer must have its own sun. As I understand this model - every observer must have it's own bending, which is - simplified - same thing...

10wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 5/9/2018 | 17:41
Ivo
Does sun rays have different bending for any observer, for example due east from my position (45°N) and different bending, for example due south?

Yes indeed. The sun rays have to be different for each observer, depending on his location and time of day.

Ivo
and what happens during the day on equinox?

Equinox are the only 2 days of the year where the sun raises exactly due east and sets exactly due west for each observer on the earth. Every other day of the year the sun raises and sets on an slightly other direction. This is due to the fact that the earth axis is tilted with respect to the plane of earth's orbit around the sun. The axis of the earth points always to the same direction on the sky, Polaris.

https://en.wikipedia.org/wiki/Equinox#/media/File:AxialTiltObliquity.png

On summer at the northern hemisphere the earth is in such a position on the orbit, that the tilt is towards sun. This causes that the terminator line (line of dawn) on the northern hemisphere reaches over the northpole, so the days on the northern hemisphere are longer than 12 hours and the sun raises and sets in a more northern direction than exactly east/west. 6 months later the earth is at the opposite position on the orbit, so that the northern part of the axis is tiltet away from the sun, producing winter on the northern hemisphere. It is the opposite on the southern hemisphere.

https://en.wikipedia.org/wiki/Equinox#/media/File:North_season.jpg

At Equinox the axes builds a right angle with the line earth-sun. The terminator line passes through the north and south pole. This causes that the day on every point of the earth is exactly 12 hours long and the sun raises exactly due east and sets due west everywhere.

https://simple.wikipedia.org/wiki/Equinox#/media/File:Earth-lighting-equinox_EN.png

See also Wikipedia: Equinox

Here is how the light rays of the sun are bent at a certain time to the observers on the Flat Earth:

http://walter.bislins.ch/bloge/index.asp?page=Flat+Earth+Dome+Model&state=--9-378.5-9-31-21-2

To rotate the model use the red sliders. To change the time or date use the black sliders. To change the location of the observer use the green sliders. Click the buttons above the animation to watch some demos with explanations.

11daznez 6/7/2018 | 06:06

Jason, you are arguing with someone who is still naive enough to believe the moon landings were real, and doesn't know how wikipedia actually works; presumably he has never read orwell either, and still believes saddam had weapons of mass destruction..

you do however have some genuinely fascinating ideas regarding the upper barrier of our fixed plane, electromagnetism and the bending of light.

feel free to get in touch, or provide links to research/ ideas - daz@uymail.com

regards,
darren nesbit.

12wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 6/7/2018 | 15:51

@daznet, I know how Wikipedia works. It is self correcting. If you post BS it will be deleted within seconds. If you post something you can't back up, it will be deleted. Wikipedia does back up each fact with references to accepted, peer-reviewed and original sources.

Of course you have to deny Wikipedia and all facts, because it destroys your conspiracy believes. If you don't trust Wikipedia, why do you not research the original papers, reports, images, videos ect.? Can you back-up anything you tell at all? Youtube videos are not peer-reviewed and scientifically accepted sources of information. They may at best be used to illustrate something like graphics do in documents.

How do you know what I have read and what I believe about Saddam and what has this to do with the shape of the earth? No other arguments?

Concerning the Moon Landings and the Apollo missions: I will present you some facts you probably don't know exist. And then we will see whether you are indoctrinated enough to still believe what some conspiracy people did tell you on youtube from ignorance or whether you are able to read the overwhelming peer reviewed and confirmed facts presented.

All about the Apollo Program and the Moon Landings

The Apollo program is fully documented in great detail. We can find thousands of scans of raw images (pre-Photoshop, no CGI), hundreds of images of the earth, hundreds of videos and audio transmissions and their transcriptions, tens of thousands of pages of mission reports, science reports, technical debriefings, scientific results, sample cataloges, and many more on the National Space Science Data Center NSSDC (links below).

It is cheaper to go to and land on the moon than to fake all this.
Why would NASA produce and fake all this documents? To deceive people that are too lazy to read it anyway?

Moon Rocks

The rockets were built and tracked by astronomers to the moon and back all over the world. The Apollo missions brought back 2415 moon rock samples weighing 381 kg that were analyzed by thousands of scientists worldwide (more Infos here). This samples are very different than rock from earth and can not be faked, because it would take hundreds of years to imitate such rocks.

Retroreflectors

There are still the 5 Retroreflectors on the moon, 3 from Apollo missions. We direct lasers to the reflectors at least once a day since more than 48 years to measure the distance to the moon to better than cm accuracy. Between 1969 and 2017 there are reported 25,689 lunar laser ranging (LLR) measurements. You can find detailed infos about LLR and download the data from Paris Observatory Lunar Analysis Center. More Infos on Wikipedia Lunar Laser Ranging experiment.

Images from Chinas Moon Lander Chang'e 3 and the Lunar Reconnaissance Orbiter

There is currently China's lander Chang'e-3 on the moon and the images sent to earth from the surface look exactly like the images from the Apollo missions: no stars, no blast crater, same color and lighting. We have space probes orbiting the moon right now, eg. the Lunar Reconnaissance Orbiter LRO, that sent back images of the Apollo landing sites and we can even see the phaths of the astronauts that match exactly the reports.

LROC Featured Lunar Sites: Some of the most requested LROC images; from newly discovered lunar features to the closest images of the Apollo landing sites since the astronauts left:

Here are the links to the archive of the Apollo missions:

Apollo Lunar Surface Journal from the National Space Science Data Center NSSDC

All mission reports, images, videos, audios, transcripts, science reports, sample cataloges, technical debriefings, biomedical results, scientific results and many, many more:

Preliminary Science Reports Apollo 11-17

Master Catalog

The NSSDC Master Catalog is available for the queries pertaining to information about data that are archived at the NSSDC as well as for supplemental information:

Apollo 11 transcript whole flight

13Aaron Marks 6/24/2018 | 14:54

Fantastic job with this model. Keep up the fight for knowledge, Walter. It's astounding how people are refusing to accept even basic explanations of physics, or resort to conspiracism and accusations of "faking" of evidence when their model fails. The truth will win out eventually.

14Jerome 6/25/2018 | 15:37

Hi Jason,
if the sun is the intersection of electromagnetic fields and not a ball.
this raises multiple questions:
what is the moon?
and how the moon can cause eclipses? (and planets passing between us and the sun)
what is the distance of the sun?
what are we observing with telescopes when we see the sun?
what are all these details which can see?
why its a perfect sphere?
why we dont see any electromagnetic artefacts around the sun?
whats the strengh of this electromagnetic field? and why we can measure/detect it?
whats generating this field?

for the level experiment…
try this:
get your phone with the level app.
put it on top of a ball
rotate the ball without moving the phone
did you see the level changing?
that's exactly what's the earth is doing, no matter where you are on the globe the down is ALWAYS pointing to the center of earth.

15Michael Dunlavey 7/1/2018 | 02:28

Nice work!
BTW, we just spent a week in Montreux, where there's a beautiful armillary sundial: https://imgur.com/PWsCs7I
It would be nice if any flat-earth fan can explain how those work.

16Alfredo Jaramillo 7/2/2018 | 21:09

hello, I loved the work you did here.

I'm organizing the next experiment

https://docs.google.com/document/d/1_MDbyJOily5eKbbKrW9ob7VZlfL4KNf4RSr1F4rDRtA/edit

would you helps us showing the results? Maybe programming an applet like the one you show in this article?

You can also steal the idea exposed in the link :)

greetings

17cearnicus 7/8/2018 | 18:58

Holy carp that's damn impressive. You really make the best models :)

I think there might be a way to get the southern celestial pole in there as well. Instead of a simple dome, use a dome with a hanging part at the center. It's ... difficult to put it in text, but a bit like a uvula in the mouth. The neck of this uvula is on the rotational axis. Make the neck 0 in size and place the southern celestial pole there. The rest of the uvula is then the celestial sphere. The center is the celestial equator and the lowest part the celestial north pole.

This arrangement allows you to have two parts of the sky on the rotational axis. I haven't checked it fully, but I think you should be able to make a mapping to the celestial sphere. Sure, it'll take even more ridiculous light bending, but since you need that anyway for the other stuff, you might as well go one step further, right?

And yes, I know that what I'm suggesting is utterly ludicrous and convoluted. But so is the rest of the flat earth model.

EDIT: alternatively, just lose the dome entirely and place a sphere anywhere on the rotational axis. This is literally the celestial sphere, only inverted. The bottom is the 'north pole' and the top maps to the 'south pole'. Since we're using unrealistic light-bending anyway, the specific geometry of the thing even doesn't matter.

18David Ridlen 8/20/2018 | 06:20

https://www.youtube.com/watch?v=uexZbunD7Jg

19Tureco 9/8/2018 | 04:03

Hello,

Questions:

1. Why did you choose to assume light bends through the dome when we do not know yet what the dome is made of, its actual size and shape, we don't know for sure size of sun and how its light works...remember flat earth is pretty new still and needs time to make more observations of our flat world to get a working model.

how can you still believe in the heliocentric model considering all its obvious flaws as well as all the fake nasa "evidence".

20wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 9/8/2018 | 19:34

Tureco, are you serious? I am sure the heliocentric model is right, because it explains all observations and makes predictions that match reality.

All calculations I had to make to get this Flat Earth Dome model to conform with some observations are based on the heliocentric model, because it can't be derived from any imaginable flat earth model! The only way the flat earth model could at least partially explain some observations is by bending the light as seen in the model above. Read Problems with the Model.

Tureco: we don't know for sure size of sun and how its light works

Of course we know the size and distance of the sun. We have measured it in multiple ways and all kind of measurements get the same answers.

And we know exactly how light and other electromagentic waves work. Your smart phone is based on this knowledge: camera lenses, liquid cristal displays, radio transmitter and receivers. I know this, because I'm an electric engineer that has to apply this knowledge to build such devices.

We have technology to see the earth from space since more than 50 years, not only NASA by the way. We have non NASA satellites in space that send full scale color images of the earth in resultions up to 11000x11000 pixels every 10 minutes. And this images match the images from NASA perfectly. You can download the raw images for free. You can build your own receiver to receive their date and show it on your computer. The satellites are real and they are in space!

How to Receive Beautiful Images of the Earth Directly From Space | GOES-15,16,17 and Himawari 8 HRIT

The NASA Apollo program is fully documented in great detail. We can find thousands of scans of raw images (pre-Photoshop, no CGI), hundreds of images of the earth, hundreds of videos and audio transmissions and their transcriptions, tens of thousands of pages of mission reports, science reports, technical debriefings, scientific results, sample cataloges, and many more on the National Space Science Data Center NSSDC (links below).

Why would NASA produce and fake all this documents? To deceive people that are too lazy to read it anyway?

Today with the aid of satellites we can easily measure the size of the whole earth with cm accuracy! From where do you think we get the data for all maps we have to navigate around the world? We navigate around the globe with navigation systems that depend on gravity and the rotation of the earth to be able to find true north and the latitude without reference to the outside. The flat earth gyro experiments are a joke. You cannot measure the rotation of the earth with such toys. Look here how real gyros are built and compare:

Gyrocompass Theory and Operation - madden-maritime.com

Flat earth has no model yet. Why? It's not because it is still new. The flat earth idea is thousands of years old. They have access to the same technology but never got any model? Nothing that can be tested? Only so called experiments, that prove nothing, because we can not make any predictions without math models that could be verified. That's the reason why flat earther say they need no model.

The Scientific by Method-Richard Feynman:

Flat earth fails in every single aspect.

21Argy Bargy 10/29/2018 | 02:41

This article is very interesting, but immediately flawed! You are a shill, oh yes you are!

Orlando Ferguson's 'Square And Stationary Earth 1893, which I have to presume you are familiar with, would remedy your model's major concession - that of needing to stretch light in Antarctica.

It's because I know that you you are familiar with Ferguson's depiction, that I know that you are a shill. Only a shill would go to such laborious effort. Equally, if you're familiar with Ferguson's work, I say that only a moron would go to such laborious effort, without researching past efforts.

You have contributed, though. You have just confirmed to FETs that, the only thing holding back their theory, is dealing with sunlight at the poles. Thanks for the tip but, as I say, Ferguson has it covered (apparently).

22wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 10/29/2018 | 19:45

No other model than the Heliocentric model with an orbiting rotating globe with an orbiting moon can accurately explain and predict all observations. You can bend the Heliocentirc model as much as you want to get some sort of almost working flat earth model, like this Flat Earth Dome model, but it never can explain all observations accurately. Only one geometry can do it: the Heliocentric model.

23Karissa Best 10/29/2018 | 23:39

In regards to your comment,
"Gravity is an acceleration, beacuse all objects in free fall are accelerated by the same rate of 9.81 m/s2 towards the center of the earth. How does a MEMS measure acceleration?"
That's not true. 9.8ms is actually not consistently the same rate. It is NOT a constant force as the mainstream teaches us. They even admit this. It is only that rate in the lower part of the atmosphere. That's because it's due to air density rather than gravity.... as we ascend into the lesser density sky it actually becomes faster.
I've made a video on this, please see:
(url)

External Links and Pictures in the comment are disabled and are not displayed!
The Admin activates the Links and Pictures after checking.
24wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 10/30/2018 | 01:51

Gravity, Buoyancy and Air Resistance (Drag)

@Karissa Best. I am an engineer and as such have to know about gravity, air resistence and buoyancy. You have mixed up everything:

Karissa: 9.8ms is actually not consistently the same rate. It is NOT a constant force as the mainstream teaches us. They even admit this.

Mainstream science does not teach that the gravitational acceleration is constant. It depends on the distance to the center of the earth, but in the atmosphere it can be treated as practically the same. It reduces only a fraction at altitudes of airplanes or on mountains. See below.

Karissa: It is only that rate in the lower part of the atmosphere. That's because it's due to air density rather than gravity.

Air density has no influence on gravity. But you are subject to both forces, gravity and buoyancy. The force of gravity acts to the center of the earth, the buoyant force acts in the opposite direction, because it is a consequence of gravity. The buoyant force is dependent on the density of the medium. The density of air decreases with altitude, so the buoyant force decreases with altitude too. Gravity also decreases with altitude.

So what is the net effect? The buoyant force of normal bodies (not helium balloons) compared with the force of gravity is negligible. For example the buoyant force of a man with a mass of 80 kg, which corresponds roughly to 80 litres = 0.08 m3 volume, is only:

(1)
F_b = \rho \cdot V \cdot g = 0{.}942\ \mathrm{N}
where'
F_b ' =' 'buoyant force acting in the opposite direction of gravity
\rho ' =' '1.2 kg/m3 = density of air at sea level
V ' =' '0.08 m3 = volume of the man
g ' =' '9.81 m/s2 = gravitational acceleration at sea level

The gravitational force on the other hand for this man is about 785 N. So buoyancy can be neglegted here. As you get higher the air density decreases very fast, while the gravitational acceleration decreases only slowly. So buoyancy gets even more neglegtable with altitude.

For a lifting helium balloon the buoyant force is greater than the force of gravity, because the mass of the helium of a certain volume is less than the mass of the displaced air of the same volume. But gravity is still acting on a helium balloon, because the mass of helium and the balloon with payload is not zero! But the net force of gravity and buoynacy is pointing up if buoyancy is greater than gravity, so the balloon raises.

The balance between buoyancy and gravity is the reason why a certain helium balloon can only reach a certain altitude, where the weight of the displaced thinner air is exactly the same as the weight of the whole ballon, inclusive helium, hull and payload.

So to calculate how much helium you need for a payload to reach a desired altitude, you have to calculate the force of gravity and the buoyant force for that altitude. From that you can calculate the volume of the balloon needed. So for such calculations you have to take into account gravity and buoyancy is also dependent on gravity.

Can you do such calculations without gravity? I guess not.

Karissa: as we ascend into the lesser density sky it [gravity?] actually becomes faster.

What? Makes no sense.

Below is how I calculate the forces. Note: you are always subject to all forces at the same time. So you have generally to take into account all forces, add them up and the resultant force tells you, how much you are accelerated. But in many cases some of the forces are neglegtable compared to the other forces or cancel each other. But nonetheless they are all there together with some contact forces, eg. from the ground. If you are standing on the ground, the gravitational force and the buoyant force are canceled by the force the ground is exerting on you, so your acceleration is zero.

Gravity

The acceleration due to gravity is dependent on the distance to the center of the earth:

(2)
g = { G \cdot M \over d^2 }
where'
g ' =' 'gravitational acceleration at distance d
G ' =' '6.674 × 10−11 m3/kg/s2 = Gravitational constant
M ' =' '5.972 × 1024 kg = mass of the earth
d ' =' 'distance of the object from the center of the earth

If you plug in the distance of the surface of the earth into the equation you get the well known acceleration at the surface of the earth:

(3)
g = { 6{.}674 \times 10^{-11}\ \mathrm{m}^{3}/\mathrm{kg}/\mathrm{s}^{2} \cdot 5{.}972 \times 10^{24}\ \mathrm{kg} \over (6371\ \mathrm{km})^2 } = 9{.}82\ \mathrm{m}/\mathrm{s}^{2}

At a distance of 6371 km from the surface of the earth you are double the distance from the center of the earth and the acceleration at that altitude is still 1/4 of that: g = 2.45 m/s2.

As long as you stay in the atmosphere until about 100 km the reduction of the gravitational acceleration is very small: g(100km) = 9.52 m/s2.

Air Resistance (Drag)

The air resistance on a free fall is dependent on the density of the air at the current altitude, the shape of the falling body and the current speed:

(4)
F_D = {1 \over 2} \cdot \rho \cdot v^2 \cdot c_D \cdot A
where'
F_D ' =' 'drag force = air resistance
\rho ' =' 'density of the air at the current altitude
v ' =' 'free fall speed
c_D ' =' 'drag coefficient, dependent on the shape of the falling object
A ' =' 'cross section area of the falling object as seen from above

If the air resistance gets equal (but opposite) to the force of gravity Fg = m · g, where m is the mass of the falling object, you have reached terminal velocity. At terminal velocity the force of gravity and the air resistance force cancel each other so the net force is zero. Zero net force means there is no more acceleration, so your speed stays constant, hence terminal velocity.

Because the air resistence is dependent on the air density at the current altitude, the terminal velocity is also dependent on the altitude.

Because the air density at high altitude is less than at low altitude, you have to reach a higher velocity at high altitude to get a drag force equal to the gravitational force. So in high altitude the terminal velocity is much higher than at low altitude.

Buoyancy

Buoyancy is a force too. It is equal to the weight of a volume of the medium which is displaced by an object in this medium:

(5)
F_b = \rho \cdot V \cdot g
where'
F_b ' =' 'buoyant force acting opposite to gravity
\rho ' =' 'density of the medium, eg. air or water
V ' =' 'volume of the submerged body
g ' =' 'gravitational acceleration at the location of the body

Note that the gravitational acceleration plays a role here! Without gravity there is no buoyancy, as experiments in zero gravity show, where liquids with different densities to not separate anymore!

25Karissa best 11/3/2018 | 23:26

Respectfully,
Gravity is unable to be proven (it doesn't exist is why),
You suggest density has nothing to do with the speed of acceleration, but it has EVERYTHING to do with it. How can the density of the air NOT affect how fast things fall through it.
You'll find if you exclude "gravity" out of everything.... that it suddenly makes so much more sense.

26wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 11/4/2018 | 17:04

Karissa best: You suggest density has nothing to do with the speed of acceleration, but it has EVERYTHING to do with it. How can the density of the air NOT affect how fast things fall through it.

I said "Air density has no influence on gravity". Gravity is independent of density. BUT: falling things are slowed down by friction due to falling through a medium of a certain density. See Air Resistance (Drag).

You did not understand that there always are multiple forces acting on anything in the atmosphere at the same time: Gravity, Drag if the object moves through the medium, and Buoyancy, and if the object has contact with other things like the ground there are also contact forces. All forces together determine how an object is accelerated. From the acceleration we can calculate its trajectory.

All about Gravity

Karissa best: Gravity is unable to be proven (it doesn't exist is why)

Wrong, Gravity is real. An engineer has to deal with it in many, many calculations. There are many devices that only function because of gravity. How do we know that it's gravity what causes their functioning?

Observing Gravity

How can we measure and observe the properties of gravity?

We can observe that all masses on earth are accelerated the same amount per time unit with g = 9.81 m/s2 on the surface of the earth, if we can neglegt the drag and buoyant force.

We can measure that gravity gets weaker with altitude exactly as predicted by Newtons theory of gravitation.

We can observe the orbits of planets and moons, due to gravity. This orbits can be computed with Newtons law of gravity.

We can observe the period of a pendulum which is only dependent on the length of the pendulum and gravity g.

We can measure the force of gravity with a scale. Because gravity is a force, it can be summed with other forces. Eg, on the spinning earth there is a latitude dependent centrifugal force acting perpendicular away from earth's axis of rotation. What we measure with a scale on any point on earth is the vector sum of gravity and the centrifugal force. That means a mass has different weight depending on the latitude. This is the reason why scales have to be calibrated for different latitudes. See my Earth Gravity Calculator.

Devices depending on Gravity

We know exactly all properties of gravity. Its properties are the thing many devices are based on. No density or buoyancy can explain it. They are completely other things and have completely other math to be explained than gravity. You can't replace gravity with density/buoyancy/weight. It simply does not work in reality. I could prove it to you, but math and physics is certainly not in reach of your understanding.

All inertial navigation systems (based on gyroscopes) rely on gravity and the rotation of the earth to sense the directionn to true north and calculate the latidude without any reference to the outside. Their mounting platforms are mounted in gimbals and have to incorporate Schuler tuning, so they always are parallel to the surface of the earth. Like a pendulum, Schuler tuning is only dependent on gravity and the radius of the earth.

https://en.wikipedia.org/wiki/Schuler_tuning

Check this out. Gyrocompass Theory and Operation - madden-maritime.com

You see, there are devices that rely on the fact that gravity exists and that the earth is a rotating globe. You can't simply deny it. No airplane or rocket could navigate without such devices. Fact!

Measuring Gravity

There are many ways to measure gravity directly: Cavendish experiment (I personally executed this experiment in physics class to determine the gravitational constant G 35 years ago). This experiment is repeated uncountable times each year as part of physics lectures. Orbital periods depend on the gravity of the central body. You can measure gravity with a simple pendulum. The period of a pendulum depends on gravity and the length of the pendulum only. So if you know the length of a pendulum and the period you can compute gravity g. You can't calculate the desired path of any rocket or ballistic object without knowing exactly how gravity works.

Beside the Cavendish experiment we can measure the gravitational force that a mountain or hill exerts on something beside it. This was done e.g. with the Schiehallion experiment:

https://en.m.wikipedia.org/wiki/Schiehallion_experiment

There are a bunch of methods to measure the gravitational constant G = 6.67 × 10−11 m3/kg/s2:

  • torsion balance, dynamic mode
  • torsion balance, dynamic compensation
  • suspended body, displacement
  • stationary body, weight change
  • attraction between a cloud of cold rubidium atoms and tungsten weights
  • Schiehallion experiment, indirect measurement
  • atom interferometry
  • pendulum period change due to testmass
  • beam balance
  • Free-fall absolute gravimeters and gradiometers

More Infos:

Properties of Gravity

Properties of gravity predicted from General Relativity and confirmed by uncountable experiments are:

  • The Equivalence principle: observation that the gravitational "force" as experienced standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in an accelerated frame of reference far from any gravity. Equivalence principle (Wikipedia)
  • Bending of light and other EM waves due to spacetime curvature near massive objects like stars, black holes and galaxies
  • Gravitational Time dilation, taken into account in GPS navigation systems
  • Gravitational waves, see LIGO
  • Black Holes
  • Frame-dragging effect, the amount by which the rotating Earth drags its local spacetime around with it: General Relativity and Gravity Probe-B - Barry Muhlfelder (SETI Talks) (YouTube)
    Official site of Gravity Probe B with all data: einstein.stanford.edu

See also:

Scientific Theory of Gravity

Newtons and Einstein's scientific theories of gravity are one of the most tested theories. They never failed in experiments accessable to us. Note: a scientific theory is not a guess or idea or hypothesis.

A scientific theory is an explanation of an aspect of the natural world that can be repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results. Established scientific theories like Newtons and Einsteins theory of Gravitation, have withstood rigorous scrutiny and embody scientific knowledge. Source: Scientific theory on Wikipedia

If someone wants to replace the theory of gravity, he has to come up with a theory, that has the same power of prediction of what the current theory has and it must be simpler or explain things we currently can't explain yet.

27Ada 11/7/2018 | 17:00

One problem I've noticed with all the flat earth models is the sun and moon are shown as travelling in a circle above the sky. They actually travel in an arc across the sky with a different starting point in each season (example: directly east to west, southeast to southwest, exactly east to west, northeast to northwest and back to directly east to west. You can see this in the annual SkyWatch magazine.

28Nicolas 11/9/2018 | 10:34

All the observations over selestial sphere tell us that our earth is spherical - different rotation of it in nothern and southern hemispheres, corresponding of local coordinates and selestial ones and so on. Then, we can observe the bending of geoide not only from seashore (lowering of visible horizon), but in any place of earth with geodesical equipment. So, any possibility of Flat Earth can base exceptionally on these three elephants:
1elephant: there is Holographic Dome over this hypothetical Flat Earth, that is modelling spherical illusion. And there are no not only stars, but Sun and Moon too, as real physical objects. And this Dome could be a huge living organism, and it have grown itself, and nobody built it. And any conspiracy theory is not needed;
2elephant: our world map is wrong in high southern latitudes at least. But if we have Holographic Dome, we may suspect, that it cheats us sometimes. Our maps was made with close binding to selestial coordinates, esp. in Southern Ocean and in ocean at all. So, it could be possible,
3elephant: there is negative refraction in our atmosphere. In other words, light beams are bending upword, and making illusion than geoide is bending downword, and horizon go down. Physics tells us about positive refraction, but who knows?
And these three elephants are standing on the turtle with Socrates head, saying: I know that I know nothing, what really spacecraft and gps and many other things are

29dharma 11/12/2018 | 09:11

It would be great of you made a cosmocentric model

30giovani 11/12/2018 | 11:18

i wonder why flat earthers remains so retarded

31John 11/13/2018 | 23:25

I don't understand what your point is, Karissa Best. Gravity has been proven several times over. In addition, excluding gravity would result in floating, like we would experience in space.
Thanks wabis. I understand the flaws in flat earthers theories even clearer now.

32cearnicus 11/28/2018 | 14:36

Re gravity: it may also be instructive to look at how one can derive the basic form of Newton's law of gravitation from the observations.

In 2d, circular motion is given by (x, y) = (r \cdot cos( \omega t ), r \cdot sin( \omega t) ), with r the circle's radius and ω the angular velocity. To get the velocity and acceleration you have to differentiate once and twice with respect to time. If you look at the magnitude of those, you have

v = \omega r (= speed along the circle)

a = \omega^2 r = v^2 / r ( = centripetal acceleration)

Note: this stuff is simply part of motion along a circular path. This has nothing to do with gravity (yet), this is what it means to move in a circle. In terms of forces, by definition \vec{F} = m \cdot \vec{a}, meaning that any force that points directly inward and has the same magnitude at constant r results in a circular path.

Kepler's laws provide an accurate description of planetary motion. They are essentially summaries of the observations. Simplifying things to circles, his third law states that the radius cubed is proportional to the period squared: r^3 \propto T^2. Again, no magic; this is what the observations tell us.

So, let's rewrite that so that it fits the formula for the centripetal force.

\begin{eqnarray} T^2 \propto r^3 \\ \left( \frac{2 \pi}{T}\right)^2 \propto \frac{1}{r^3} \\ \left( \frac{2 \pi}{T}\right)^2 r \propto \frac{1}{r^2} \\ \omega^2 r \propto \frac{1}{r^2} \\ a_{centripetal} \propto \frac{1}{r^2} \end{eqnarray}

In other words, the pattern we see in planetary motion can be explained by a force that's inversely proportional to the distance squared between the planet and the sun. And, again, since \vec{F} = m \cdot \vec{a}, we also know that this force is proportional to the mass m_p of the planet:

F_p \propto m_p / r^2.

Now, technically it's not just simply F_1. It's actually the force of one object (the sun S) acting on another (a planet P). So really we have F_{s\rightarrow p} \propto m_p / r^2.Finally, from Newton's third law we know action = -reaction, so F_{s \rightarrow p} = -F_{p \rightarrow s}. For that we have the something similar: F_{p \rightarrow s} \propto m_s \ r^2.

Putting all of this together, we get:

F \propto \frac{m_{s} \cdot m_{p}}{r_{sp}^2}

Again, all I've done here is distill the observed paths of planets into a formula of a force. And, oh hey, looks suspiciously like Newton's Law of gravity. Note that right now I've only looked at circles and magnitudes, but if you generalize it to vectors and ellipses, you'll find that the law of gravity correctly describes all of Kepler's laws (see its wikipedia page for details). It also correctly predicts that the falling acceleration on earth should decrease with altitude, the atmospheric pressure profile, buoyancy, period-vs-distance relation between the moons of the solar system, etc, etc.

So yes, a force of this form is required to accurately describe nature. We call this force "gravity". Flat earthers can whine about it all they like, but this is what the observations tell us.

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