Concave Earth Theory

Direct evidence

1. Old maps
2. Modern maps
3. 19th century balloon observations
4. Tamarack mines
5. Laser between two posts
6. Rectilineator

Indirect evidence

7. Huge horizons
8. Binocular effect
9. Bendy light
10. Overall conclusion
+++

Indirect evidence

7. Huge horizons

On a convex Earth there is a Pythagorean way of calculating the horizon. It is based on two assumptions: 1. The observer rests on a convex sphere, and 2. Light moves in straight lines. Both convex heliocentrists and geocentrists believe these tenets to be near practically true. The Earth is an oblate spheroid which for all intents and purposes is practically the same as a sphere (maximum 0.34% difference); and light does travel in straight lines except for minor deviations when entering air of different densities, temperature and/or humidity – refraction. (There are big problems with this model when trying to triangulate the Sun, but that is another topic.)

dip1
The Pythagorean equation shows us that the convex ball Earth is stopping light from a distant object hitting our eyes at “OG” distance away.

As long as the height of the observer (and object observed) is less than the radius of the Earth (which it always is) then the simple equation is OG = square root of 2 x (R x h). The radius of the Earth is about 6378 km or 6378000 m. So, a man whose eyeballs are 1.8 m above the ground would see 6378000 x 1.8 x 2 and then square root that number = 4791.7 m or 4.79 km. Making it even easier for us are the online horizon calculators such as ringbell.co.uk and boatsafe.com. These give a horizon distance of 4.8km for an observer 1.8 m high, which is the same to one decimal place.

Refraction also needs to be added. To get an exact amount of even estimated refraction is complicated because it would depend on the height difference of the observer and object being observed. The greater the height difference, the less the refraction, as the angle of incidence of light coming from the object is more vertical. The more horizontal the path of the incoming light through a medium of different density, the more the angle of refraction.

images.duckduckgo.com
Light refracts the most at the horizontal than from vertically straight above the object.

It is even worse when trying to incorporate possible pressure (density), temperature and humidity differences; let alone try to find out any actual readings. However, the California State University states in its extra educational Youtube videos:

“Refraction or “r” is affected by temperature, atmospheric pressure and geographic location. Rather than creating a complicated equation incorporating all of these variables, engineers are safely able to assume that the refraction is roughly equal to 1/7th of the curvature area.”

+++


Refraction is roughly equal to 1/7th of the curvature area.

We just add this 1/7 to the radius of the Earth: R′ = R × 7/6. This gives us 7440km or 7440,000m for the Earth radius as the “goto” figure when calculating standard refraction into the horizon calculation. So, a 1.8m high observer can see 7440000 x 1.8 x 2 and then square root that number = 5175m, or 5.17km away, if light travels in straight lines on a convex ball. This adds about 8% on top of any figure from a horizon calculator which doesn’t calculate refraction in their equation: 5175 – 4791 = 384; 384/4791 = 0.08; 0.08 x 100 = 8%. It also agrees with the wiki figure of adding 8% for refraction to the total horizon distance calculated.

Observed contradictions
Unfortunately for these assumptions, the real world begs to differ. Below are four clear cut examples of the horizon being way further out than the above calculations allow. (There are more anecdotal ones in the horizon article).

1. YouTuber Joeseph Winthrop has conducted tests with a blue and a green laser at 20 miles distance across mostly water at night. The height of the lasers was 45 feet. The cameraman is on a pier 15 feet above the water. This is a total of 60 feet elevation which should see 4,623 (Earth radius in miles) x 0.14 (refraction) = 647.22; 647.22 + 4,623 = 5270.22 miles or 27826762 feet. 27826762 x 60 x 2 and then square root that number = 57786 feet (or 10 miles and 1661 yards). The naked eye was able to pick up both lasers, but only the camera was sensitive enough to register the blue laser pen 20 miles away. This is a fraction under twice the allowed horizon distance on a convex Earth.

blue laser horizon test
(Click to animate). “The cameraman detects two very faint direct streaks of blue light. The cameraman was otherwise able to see this laser with his eyes, and most noticeably when the direct hits were made.”

2. On the David Icke forum a previous mainstream model believer “Spock” posted that a friend of his had spotted Blackpool Pleasure Beach (amusement park) from Ainsdale beach in Merseyside 11 miles away across the beach itself (low tide – no water).

Pleasure beach distance
The distance was roughly 11 miles away.
 Rhino binoculars
Rhino Binoculars used to spot a shoreline 11 miles away.
 photo 26FBD7AA-D207-40E3-9DDE-6DE5F4FD4E90_zpsx9b5gmwf.jpg
The Blackpool shoreline across the sand is seen 11 miles away.

The camera looks to be approximately 5 foot above the flat shoreline (after the tide had gone out). Without going through the conversions and calculation for refraction we can just add 8% on to the online horizon calculator. Five foot elevation sees 2.7 miles without refraction; and 2.7 x 1.08 = 2.916 miles with refraction. The binoculars saw 3.77 times (277%) the allowed horizon distance on a convex Earth.

3. The lights of Milwaukee were seen at night across a large lake 136 km away in Grand Haven from an elevation which looks like the observer is just above the beach. The picture below isn’t clear enough to make out exactly how much of Milwaukee is visible; however, the tallest building in the city is 601 feet. Only 32 buildings are above 230 feet with all but 5 of those between 230 and 400 feet tall. The observer looks to be about 20 foot above the water level, but lets double it to be conservative. I count at least 16 bright dots on the horizon which means at least one of those 400 foot high skyscrapers was seen. This gives a grand elevation total of 440 feet (conservative – probably less). 440 feet is supposed to see 25.7 x 1.08 = 27.76 miles away, not over 84 miles distance (136 km). This is over 3 times (200%) the maximum allowed horizon.

milwaukee-lights
The Milwaukee lights seen at fantastic distances just above the beach.
Milwaukee-night-shot
The distance between Grand Haven and Milwaukee is 136km!

4. Donald Sarty found this photo of the Canadian shore from Huntington beach in Bay Village courtesy of an article on newsnet5.com dated May 7 2013.

50km shoreline
The shoreline 50 miles distant can be seen with very little elevation. Big thanks to Don for finding the video.
huntington bay horizon
The actual distance is about 50.527 miles according to daft logic from Bay Village across Lake Erie.

I do know the shoreline was visible to many this past Sunday night. Friend and fellow scientist Jay Reynolds, a professor at Cleveland State University, sent in the above pictures on Monday. This shot was taken at Huntington Beach in Bay Village, Sunday at 11 p.m. It includes a close-up shot, which clearly shows the lights of the Canadian shore. Jay measured the distance to Canada from his vantage point as about 50 miles… “Because of the curvature of the Earth, we are limited to approximately 16 miles.”

+++

We have no need to guess the elevation of the camera this time as they say they can normally see 16 miles (I assume they have included the extra 8% refraction). That is 3.12 times (212%) more than they should if the Earth were convex.

Super refraction
But wait, they say instances such as those above are caused by “super refraction”.

Unfortunately, the refraction varies considerably from day to day, and from one place to another. It is particularly variable over water: because of the high heat capacity of water, the air is nearly always at a different temperature from that of the water, so there is a thermal boundary layer, in which the temperature gradient is far from uniform.

But the structure of thermal boundary layers guarantees that there will be large variations in the gradient, even in height intervals of a few meters. This means that on two different days with the same temperatures at the eye and the water surface (and, consequently, the same dip), the distance to the horizon can be very different. In conditions that produce superior mirages, there are inversion layers in which the ray curvature exceeds that of the Earth. Then, in principle, you can see infinitely far — there really is no horizon.

+++

“You can see infinitely far” with certain thermal boundaries. Is that possible? What differences in temperature and/or humidity over the curvature drop is necessary to produce an increase in refraction between 12 and 34 times the norm in the above four examples? Has anyone done those calculations? For argument’s sake, let’s take this statement as true. The key word here is “variable“, i.e. not constant.

So the nice-looking formulae for calculating “the distance to the horizon” are really only rough approximations to the truth. You can consider them accurate to a few per cent, most of the time. But, occasionally, they will be wildly off, particularly if mirages are visible. Then it’s common to see much farther than usual — a condition known as looming.

+++

The key words are “accurate” and “most of the time“. What is looming?

The appearance above the horizon of a distant object that would normally be hidden below it. This effect is caused by unusually large terrestrial refraction, usually due to a thermal inversion. Looming is the opposite of sinking. Both are refraction phenomena, but not mirages.

+++

So super refraction is caused by temperature (thermal) inversions. This is when the air is colder at the ground or ocean than directly above it; so you have ground, then cold air, then warmer air above that.

Given enough pressure, the normal vertical temperature gradient is inverted such that the air is colder near the surface of the Earth. This can occur when, for example, a warmer, less-dense air mass moves over a cooler, denser air mass. This type of inversion occurs in the vicinity of warm fronts, and also in areas of oceanic upwelling such as along the California coast in the United States. With sufficient humidity in the cooler layer, fog is typically present below the inversion cap. An inversion is also produced whenever radiation from the surface of the earth exceeds the amount of radiation received from the sun, which commonly occurs at night, or during the winter when the angle of the sun is very low in the sky. This effect is virtually confined to land regions as the ocean retains heat far longer.

+++

So a normal inversion where the radiation from the surface of the earth exceeds the amount of radiation received from the sun is over land only; but there are cases when warm wind moves over colder air (ocean or land) – warm fronts. These are of course variable. The other cause is said to be ocean upwelling which “involves wind-driven motion of dense, cooler, and usually nutrient-rich water towards the ocean surface, replacing the warmer, usually nutrient-depleted surface water.” Wind is variable. Interestingly, inversions never occur over the oceans during winter: “there is no trace of inversion of temperature over the oceans during the colder months.”

In a nutshell, over oceans/seas thermal inversions are caused by winds, with no inversions during the winter.

Three out of the four huge horizon examples above were over water (albeit very large lakes rather than ocean). Other examples on YouTube of larger-than-normal horizons are nearly always over water. This is just to make sure elevation isn’t variable more than anything else. It is possible that the four examples above were freak events caused by inversion, and that the Blackpool Pleasure Beach shoreline is usually way below the horizon except on that day at that time.

However, a poster on the concave Earth forum, Primalredemption, claims that his huge horizon observations are always there every time he walks the beach:

I could actually see the waves crashing against the shoreline of Molokai all the way from Sandy Beach on Oahu. All the way down to the shoreline. “ho that’s Coo bra”. “Yups it’s coo” until you realize that Molokai is 66 miles away, which would put the shoreline 2900 feet beneath the horizon on a convex Earth… I was about 5 feet above sea level for the most part… I have, all my life, different times of day and seasons. It is always the same scene and therefore is not caused by atmospheric refraction. The limiting factor is the clarity of the air.

+++

The distance isn’t 66 miles, but 26 miles.

hawaii horizon
The distance between Sandy beach on Oahu to the Molokai shoreline is 26 miles.

26 miles is still a big problem. With the observer 5 feet high, the convex horizon is 2.916 miles distant (including standard refraction). Thermal inversion is ruled out, because the observations have been made at different times of day and seasons all his life, i.e. constant.

Do-it-ourselves (2)
The above example is anecdotal, however. How do we know he is really telling the truth? We don’t. We need to prove this to ourselves either way by recording a huge horizon over an ocean or sea in the colder months (winter). And to be absolutely 100% sure, it would be preferable to do this several times on different days. I would suggest a few minutes of video at around 9am, then another few minutes around 1pm and another session at around 4pm just before it gets dark (winter months). It would be uncomfortable to spend the entire day on a beach in the winter months, so splitting this up over three times throughout the day sounds doable. If you can do this on half a dozen days throughout winter or during any season in fact, then even better.

Any camera with a decent optical zoom will help to find such “huge” horizons. The Panasonic LUMIX DMC-FZ70 has a x60 magnification and allows teleconverter lenses to be added for even extra optical zoom. The FZ70 is the cheaper previous model to the the Panasonic LUMIX DMC-FZ72, but does the job we want just as well. The camera on its own costs about $250. A tripod is necessary with such magnification to stop camera shake, but otherwise the standard memory card and battery charger package is fine.

by - [-]
Price: - - - -


by - [-]
Price: - - - -

To add a teleconverter, you will need either the DMW-LA8 Lens Adapter, or this alternative. The adapter allows for the attachment of any 55mm thread teleconverter. The Panasonic seems to have cheap and powerful teleconverters, but they either don’t add much to the real magnification and/or add a lot of fog, chromatic aberration, don’t fit and above all often don’t allow the camera to focus at all. It is a much better idea to get the official telephoto lens for the DMC-FZ270/2 which is the 1.7x Panasonic DMW-LT55E – Converter. It is quite expensive ($177), but at least it should do what it says.


(Click on image). You need this 55 mm adapter to fit the 1.7x telephoto lens attachment to the Panasonic.

(Click on image). This 1.7x telephoto lens from Panasonic themselves for the Lumix will do what it says on the tin.

With the telephoto lens plus adapter the whole lot should come to about $500. This will give you an official x102 magnification. However, there is an even “zoomier” teleconverter that fits the FZ70’s 55mm adapter thread and has 2.2x magnification instead of 1.7x, and that is the $200 Raynox DCR-2025PRO High Definition 2.2x Telephoto Lens. It seems to do what it says on the tin. This will give a total of x132 optical magnification for around $550.


(Click on image). The Raynox DCR-2025PRO High Definition 2.2x Telephoto Lens seems to work and can be used with the 55mm thread adapter for the FZ70.

by - [-]
Price: - - - -

Other superzoom bridge cameras such as those from Sony and Canon have threaded lenses which means that in theory you should be able to add the Raynox 2.2 teleconverter lens with the appropriate adapter, but you will have to look into that yourself. Cost-wise, the Panasonic looks to be the most value-for-money.

If you are only going to get a super-zoom bridge camera without a telephoto lens attachment and don’t mind spending $200-$300 more, then there is always the Nikon P900 with 83x magnification and good reviews. There is an appropriate, cheap 2.2x telephoto lens attachment for the P900, but according to the reviews it is utter pap and is really only a 1.1x teleconverter. (I’d personally avoid all cheap teleconverters.)


(Click on image). The Nikon p900 has the highest stand-alone optical zoom out of all bridge cameras as of 2015.

Other frequencies of light
So far, these horizons are seen in the visible light spectrum. It seems the lower the frequency of light, the further the horizon when we look at radar and radio waves. (This is covered in more detail in the horizon article.)

Boat radar has a maximum range of 150nm (277.8 km).

A civil marine radar, for instance, may have user-selectable maximum instrumented display ranges of 72, or 96 or rarely 120 nautical miles, in accordance with international law… and maximum detection ranges of perhaps 150 nautical miles.

+++

The company Raymarine uses radar with a maximum range of 72nm or 133km. Boat radar is attached to the top decking as seen in the photos below:

open array radar on boat2
The open array radar on top of the boat looks to be about 3m? above the waterline.
open array radar on boat1
This open array radar looks to be about 6m? above the water.

Let’s take 5m as the observer height. The boat radar should have a maximum 8.64km (including normal refraction), not 277km! That is 32 (3100%) times more than it should. The radar isn’t bounced off the ionosphere as only radio waves of the AM bandwdith (3 and 30Mhz) do that – skywaves.

Marine (small-boat) radar typically operates at frequencies of 9.3 to 9.8 GHz, with most operating at 9.3 and 9.5Ghz.

+++

Radio waves are even worse. According to the ex-Navy electronics technician and the Great Yarmouth Radio Club, the horizon depends on the frequency of light:

“Given two signals of equal strength and different frequencies, lower frequencies travel further than higher ones… The ground wave follows the curvature of the Earth and its range does not depend upon the height of the antenna. However, the range does depend upon the transmitter power and also upon the operating frequency. Low frequencies travel further than high frequencies. Thus under ideal low noise conditions (noon, during winter), it is possible to communicate over distances of about 500 nautical miles at 2 MHz by using a 100 W transmitter. At 8 MHz, under the same conditions and using the same transmitter power, the maximum range is reduced to about 150 nautical miles.”

+++

Wikipedia say that radio waves bend around the Earth.

“During the day, AM signals travel by ground wave, diffracting around the curve of the earth over a distance up to a few hundred miles (or kilometers) from the signal transmitter.”

+++

Diffraction means “to break up or bend”. So their own official definition is stating that light is bending around the curve of the Earth. So how do radar and radio waves bend so much around the Earth? It isn’t temperature inversion as this is variable. Do lower frequencies refract a LOT more than visible light? Not really. This phenomenon is called dispersion, which is described more in terms of resonance between the EM wave and the refractive material:

“In a typical material like glass this will correspond to electronic excitations, and will be in the UV. As you increase the frequency of the light and get nearer to this natural frequency the magnitude of the induced oscillations increases, and hence the interaction with the light increases. This is no different from any driven harmonic oscillator. As you pass through the resonance and carry on increasing the frequency the interaction strength, and hence the refractive index will fall again.”

+++

This means there is a rise and fall of the amount of refraction as the frequency of the EM wave gets closer to the matching frequency of the refractive material. This isn’t a one direction relationship between frequency and refraction; it’s a resonance point. It gets worse. The greater horizons at lower frequency can only be explained by standard refraction in the convex Earth, if these lower frequencies bend more around the Earth. This means that lower frequencies must refract more than the higher ones to diffract around the Earth. It turns out the opposite is the case.

If we actually look at the calculations of the amount of dispersion of visible light of different frequencies through air at the same pressure and temperature, refraction increases for the higher frequencies (shorter wavelength). The refractive index is 1.000268479 for a wavelength of 1700 nm (long wavelength/lower frequency) and 1.000286581 for 300 nm (shorter wavelength/higher frequency) for example. This shows that the resonance point of air is below 300 nm, i.e. above visible light frequency.

Conclusion
In the visible light spectrum, constant huge horizons are only anecdotal. However, both radar and radio wave horizons cannot be explained on a convex ball. This evidence alone is an indication for both concave and flat Earth, giving a 75% chance that the Earth is concave.

Next: 8. Binocular Effect

882 thoughts on “Concave Earth Theory”

  1. What if the Sun comes few 100 kilometers closer to the earth during spring/summer seasons? I think that is the reason for fast increase in heat during these seasons while in winter season sun is further away causing colder temperatures. So starting at spring equinox sun begins to come closer to earth while at fall equinox, sun begins to go further away. Suns coming close will also heat up the glass sky.

    View Comment
  2. ” Neither quantitative nor qualitative researchers ever discerned an “aether drift” correlated to the (supposed) rotations of the “earth in space”. Quantitative science took this supposition as proof that no motional reference exists.”

    This isn’t correct.
    Aether Drift has been detected and this, many times.
    Even Relativists trying to prove Time Dilation, did in fact only bring the evidence for Aether Drift.

    View Comment
  3. A very good, rational article, which I did read with pleasure.
    The only missing part is the behaviour of the Aether in the Atmosphere, directly over the ground and the corresponding behaviour of Light in this Aether.
    It is very difficult to find a fixed point of reference in Nature.

    View Comment
    1. That is a bit much for me at this moment in time, although I have had an idea or two.

      I’ll tell you the idea now. I think matter is an alternating magnetic field (Gerlach experiment). Therefore, EM waves are constantly being emitted (as EM waves are being produced by alternating magnetic fields – see radio waves). I think this interpretation is foundational. Now when I spin my spoon in my coffee in the morning I produce one large vortex. When I spin it the other way (which I always do when stirring food and drink for some reason), I notice one or two or more very small vortices are formed which spin out away from the center of the liquid until they dissolve. I think these vortices are EM waves in the “aether” which are being produced all the time. It is easy to see how these are produced. When the direction of the large magnetic vortex is reversed it creates a counter flow inside the large vortex, with the original vortex flowing the other way, so little vortices are created and spin out. You will also notice turbulence in the counter spinning. This I believe is friction and creates heat. There are probably other ways to create these EM vortices without the constant counter-spinning which creates the heat side effect. I have a few ideas there too, but I will leave those to myself.

      I would be sorely tempted to use this foundation to build up matter, its properties and bonding; maybe search out other experiments which also point to this… or not.

      WH

      View Comment
  4. WH any way we can get a drawing of how you envision this? I’m with you on thinking the Moon could be the reflected sunlight. but I have so many people needing to see it.. I figure once you feel comfortable with your findings you will share your secret with the world… great work again as always…

    View Comment
    1. It is very early days yet Mac. The moon was extremely high last night. I am thinking it was having to do with direction of the plasmasphere rotation. Anyway, there is a small hiccup and that is I had envisaged the moon reflection rotation around the Sun to be going clockwise, but the sunspots (plasmasphere) travel anti-clockwise in the same direction as the Sun travels in the cavity. Otherwise it fits too well. I’ll keep observing the moon and thinking about it.

      I hope to get a fully mapped out path of the moon and cause like I did for the Sun, but that won’t be ready for at least a year I’d say, probably longer… if it all works out of course.

      EDIT: The moon reflection around the Sun must move clockwise to agree with what I am seeing so far. I wonder if there is a counter current to the sunspot rotation that causes it? I know that electrons (negative current) moves in the opposite direction to positive current in (or is that around?) a wire. If we look at standard electricity generation here:
      http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir.html

      then we can easily apply that to the Sun. The magnetic south is the geographic north pole in the earth cavity. The Sun (metal wire) in my model does move a bit perpendicular to the magnetic cavity current. The Sun moves slightly out from equinox to solstice and in towards the center from solstice to equinox (perpendicular direction). This would create a positive current moving clockwise around the Sun, whereas the electrons would move anti-clockwise. So, sunspot rotation would be caused by electron diffusion to the West (anti-clockwise), and the moon reflection rotation is caused by positive current flow to the East (clockwise).

      That would work.

      View Comment
      1. love the work brother. I was able to see the moon last night too. like you said its path was higher than the night before. i will be watching this year with you.. anything i notice i will be sure to pass along..

        View Comment
    2. I could do one without vertical inclination, but it will take too long to do and it is still in the idea stage so I won’t do it yet. Just picture the reflection from the back of the Sun (left side in this diagram), detach it from the line going through the center and move the reflection around clockwise. Full moon is how the diagram stands now. New moon is the fully detached reflection having rotated 180 degrees so that it is inline with the daylight side and so on.
      http://www.wildheretic.com/wp-content/uploads/2015/01/noon-sun-high-latitude-curve-rays-all.jpg

      View Comment
    1. I know. What is more startling is that a whole massive chunk of history has been wiped out over this time period. Only the maps remain. At least one group wants to keep us in the dark. It’s fun speculating and throwing conspiracy theories out there as to who, but I’m not bothered at the moment who the dark forces are, just that I recognize that they exist and to find out a little more of the truth.

      View Comment
      1. To be fair I guess: a skeptic would say that it’s not a concave earth model as much as a flattened sphere effect which the cartographers sketched. That is, a sphere whether concave or convex experiences distortion and this is an acknowledgement of that. For example, if you sketched a grip on an intact orange peel then flattened it against a counter – the grid lines would distort.

        Me I tend toward believing that Earth is a penal colony for rebellious souls.

        View Comment
        1. To be fair I guess: a skeptic would say that it’s not a concave earth model as much as a flattened sphere effect which the cartographers sketched. That is, a sphere whether concave or convex experiences distortion and this is an acknowledgement of that. For example, if you sketched a grip on an intact orange peel then flattened it against a counter – the grid lines would distort.

          Absolutely. But they put the lines the opposite way than history said they should have.

          Me I tend toward believing that Earth is a penal colony for rebellious souls.

          Certainly we would fit that description lol 🙂

          View Comment
  5. The moon is the back of the sun. It appears convex as the back of a mask appears convex despite it being concave. The earth is concave and along with the glass sky throws out so many possibilities it is hard for mere men to study and understand. But the moon is not there. Yet it is there because the sun and moon come together. They are created together because they are two sides of the same coin. The bible says earth is created then sun and moon. Krishna says in Vedas he is sun and moon. Together. They always go hand in hand. You can see stars through the illusion of course. It can appear that the clouds go in front of the moon even but if you go above those clouds you will see the moon still the same distance from you. It is thus like a rainbow which depends on the subjective. It is not a rock. It is not created by people other than the original person who everyone seems to have forgotten about. Your work is good.

    View Comment
    1. Thanks Chandra. It is interesting to explore alternative ideas. I think a lot of us are at the testing phase now. The debate was very 2013. It is nearly 2016 which will be the year of reality and the end of marketing. A lot of tests will be carried out over the next few years. The more the merrier.

      When/if proven that the earth is concave, well, bye bye space marketing and a lot of other things sold to us as true.

      View Comment
      1. Chandra can you tell us the source of your information? Is this something your heard about? read? or just came up with it on your own?

        Hey Wild 🙂 I was looking at the moon the other day and I had an interesting thought. If the heavens are a ball of perfectly clear water about 7000 miles in diameter over our heads and the Sun is revolving around it, then the moon could very well be the reflection of the Sun filtered through what is effectively a Crystal Ball. The massive pressures of a thousand miles deep of water could account for sonoluminescence or stars could simply be pockets of reflective material suspended in the relatively still waters of the great crystal ball, that are lit by the, always on, light bulb Sun of ours.

        Some study into the optical properties of Crystal Balls might be in order. Especially if the angle of view creates the moon phases effect and especially an eclipse effect. Now for this to work it would mean that one side of the Sun always faces down so the backside would always be shining straight into the crystal clear heavenly waters. A ‘balanced’ ball could do this. Then relative geographic positioning would give the effect of a monthly cycle, eclipses, etc.

        So the moon would be an actual object as I have always suspected, its just that its the Sun… (is a perfect irony at least lol)

        When I observe the Sun through a telescope with a Sun filter the ‘plasma’ of the light bulb blocks the surface except for the black spots which if you place a spotlight at a camera you will see everything behind the spotlight it looks black anyway (not that the moon is known for being colorful). Could it be that when filtered through thousands of miles of a ball of crystal clear water we would get the anomalous moonlight glow but have enough light filtered as to be able to see the bulb’s (sun) and filament (moon) itself?

        Also monthly lattitude shifts of the Sun could cause the Lunar ‘cycle’ effect maybe? Or perhaps the Suns distance to the crystal ball changing its focal point? Meaning the Sun would bob up and down a bit each month in relative distance to the crystal ball changing the focal length and causing a phase effect.

        A quick study into the optics of Crystal Balls should prove or disprove this pretty easy. Focal point of observer I think would be a very critical factor in the Eclipse effect. But if you can get a moon ‘phases’ effect with a ball of light through a crystal ball at the proper focal points that would be very noteworthy.

        We have to keep in mind the relatively MASSIVE size of the crystal ball and what effect that would have to the observer on the ground. This would explain why often the Sun and moon are seen in the sky at the same time as well.

        Perhaps something about the magnetic fields could explain why there would be water ‘polarized’ in such a manner so that some accumulates in the center and the rest is pushed to the edge creating a neutral zone of gravity big enough to drive a Sun through. Could it be that like magnets have north and south that gravity also has north and south?

        And on the atomic level…
        Maybe neutrons are neutral because they dont really exist per se but the ‘reflection’ made by a proton through a crystal ball in the center of every atom does exist and therefore must possess independent properties like light reflected off a filtered mirror. Is this why protons and neutrons are always equal in atoms? One is just the reflection of the other? Time being relatively much faster in smaller particles by reason of common sense, the number of protons and neutrons we think are in there could just be relative to the size and mass of the cave, sun, and crystal ball of which it is made, and that actually all atoms have 1 sun and 1 crystal ball and 1 hard metallic shell, all of varying sizes and shell properties. When they smash atoms together and destroy worlds at the Hadron Collider the pieces they think are this or that are just fragments of the electron shell that they think is tiny when it is actually a large crust containing a Sun and a Crystal ball that has been smashed to pieces in a violent fire of electrical discharge.

        Even the slightest chance of this being true is worth stopping all the smashing of atoms until we KNOW we aren’t destroying worlds as that would be very very very bad karma for us all.

        And some people think the concave Earth is a crazy idea hahaha
        I give you… Atom Worlds.

        Anyways keep it up brother the site is looking real good 🙂

        View Comment
        1. I’m interested in the micro, but haven’t looked looked much into it as of yet.

          The site could be a lot better with additions and revisions to the early stuff and adding the night sky in depth, but like Steve, I will concentrate on testing this year. It’s time.

          View Comment
      1. The theory behind the moon being the back of the Sun is that light bends by different amounts in a concave earth hence you can see both the moon and the sun at the same time sometimes. I’m going to draw a diagram I think to show this.

        View Comment
        1. Jesus reading these comments gives me a headache, you guys are all a bunch of brainless idiots, it is not hard to see what the moon is, looking at it with any telescope you can clearly see it is a rock. What about the planets and their moons I can watch orbit them? I am curious as to what you think those are?

          View Comment
    2. Nice try genius but for a good part of the month you can actually see the moon in the sky during the day. This means the moon and Sun at the same time as two separate bodies in different parts of the sky.

      View Comment
      1. Yes, we know all this. 🙂 Keep following the discussions.

        This means the moon and Sun at the same time as two separate bodies in different parts of the sky.

        No, it doesn’t because light bends in the cavity at different rates at different times during the day and night. Hence when the observations of the moon are complete this year, I’ll take the proposition that the moon and sun are the same thing and see how the model works. I already know how it works for tonight’s and last night’s showing.

        Nobody knows for sure, it’s only an idea.

        Tonight, the moon (at just after 6pm) is following the path of the Sun at the equinoxes (very, very roughly). The moon is just behind where I saw the sun today but about twice as high up off the ground.

        View Comment
        1. if I get a clear sky id love to help observe in ohio for you… just not sure how you are conducting experiment? some people just cant see how the CONCAVE model works… but once you see it YOU GET IT!!!!

          View Comment
          1. Exactly. Once you get it, there is no going back. I’ve figured out the moon path and phases as well. It’s reflected sunlight being pushed around by the Sun’s plasmasphere. Sunlight during the day isn’t pushed around the plasmasphere because it is the main direct light and is far too strong. A small portion of the Sun’s light is pushed around however, and this is reflected off both the front and back of the Sun which forms the “moon” and its paths and phases. A new moon is invisible because the reflection is once again inline with the daylight Sun (front of the Sun). Get it? I felt good after finally getting it. I could have only done it by observing the moon in the sky and then predicting where and when it it was going to show up next. I feel I am 100% right on this.

            So if nothing else good came about from getting that x60 zoom bridge camera, then the money was very well spent. I couldn’t figure out the moon looking at it on paper.

            View Comment
        2. I think you are going to find that the sun and moon are one and the same, just different effects. What I can’t figure out is which one is creating the other one or if they are both created by something else altogether. Did you ever see the video Crrow777 posted showing the anomaly on the moon within the same 24 hours there was a sun flare on the sun in the exact spot? I’ll see if I can find both videos…

          View Comment
          1. Thanks for that. it’s great videoing the moon yourself. I’ve found it becomes a lot easier to visualize how it would look in a certain model. The moon I have been recording looks just like the video above except his camera looks to have a much higher exposure in low light. I can’t see stars in my mini sensor high zoom bridge camera. My take on that light point at the edge of the moon is just that it is part of the of the moon being lit up by the sun. It’s really a continuation from the crescent even though there is a non-lit part before it to the lit crescent part.

            When I videoed the moon a a few days ago during the end of its last quarter phase, for two or three days in a row there was always a “star” at the same distance from the moon to the right. It followed the moon around. The star was still visible when the moon disappeared when the Sun came up. I’m guessing this “star” was Venus.

            View Comment
          2. I’ve had a bit of a breakthrough regarding the moon and how it reflects light in terms of the moon being reflected light from the back of the Sun. I was mapping out the reflected beams in my head to what I have been seeing in the sky the last week or so. I have found that the bend of the light rays match the same as the bend of the Sun’s rays except that these moon rays (reflected sun rays) are revolving around the Sun itself, NOT the Earth cavity. It also explains the phases perfectly.

            The only thing I know to immediately revolve around the Sun is the Sun’s plasmasphere. So I looked up its revolution time and lo and behold sunspot rotation was first discovered in the 19th century to take 27.28 days. http://sohowww.nascom.nasa.gov/explore/lessons/diffrot9_12.html

            Guess how fast the moon is said to take one revolution around the Earth? 27.3 days – https://answers.yahoo.com/question/index?qid=20061210034237AAdG0wW

            Here is another:
            “As any amateur who has studied sunspots drift slowly across the sun from day to day though special filters would know, the Sun’s rotation period is about 27 days.” – http://www.smh.com.au/news/Big-Questions/Does-the-sun-revolve-on-an-axis-or-is-it-stationary-in-the-middleof-the-solar-system/2005/06/10/1118347582434.html

            I know sunspot rotation is supposed to be between 25 and 35 days from equator to pole, but actual observed times seem to be about 27 days.

            That is 90% of the moon puzzle solved right there. Just the moon and sun eclipses to go.

            View Comment
    1. Hi Space, long time no hear.

      We don’t know, Space, if moonlight bends or not as we haven’t tested it. Could it be tested like the bendy light experiment, but at night with no light-bulbs at the sighting board? The moon would have to be very bright. Would it be bright enough? I’m not sure. Interesting experiment if it could be done. I assume it bends like sunlight or the light-bulbs or any other non-coherent diffuse light, but less, because it is night time.

      View Comment
      1. Hi WH,

        Did you noticed, that when moon sets, it not changes color. Sun when sets, becomes orange, red, various colors, atmospheric effects, bending light, horizon effect.
        Moonlight seems not affect atmosphere. Need to check somewhere by seashore, disappearing ship effect. Of course, must be full moon. Some photo cameras can make photos at night.

        View Comment
        1. Great observation. Never thought of that! I’d like to check that myself. It has to be because at night light bends the least, hence the lack of distortion which they say is down to atmospheric refraction, but must really be down to the electric effect on the bend of light. Otherwise the moon would show the same distortion as the Sun on the horizon, but doesn’t.

          You’ve got something there.

          View Comment
          1. Moon disappeared today Don, this morning. It just fades out. I’m going to capture it tomorrow morning on camera. I never thought to film it this morning.

            View Comment

Comments are closed.