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##### Flat Earth Theory / The FE model can't have Moon during night.
« on: July 07, 2014, 11:47:34 AM »
(Message edited to correct some typos and fix a pair of issues)

The FE model uses a light hypothesys that allows us demonstrate that the Moon can't be seen during night.

We will cover several aspects from the FE model and use several of their hypothesys.

Lets knee deep into the thing:

First, I will review what FE model says about the Day and Night cycles, and how they work, if their earth was flat:

Quote
(http://wiki.tfes.org/FAQ#How_do_you_explain_day.2Fnight_cycles_and_seasons.3F)
Day and night cycles are easily explained on a flat earth. The sun moves in circles around the North Pole. When it is over your head, it's day. When it's not, it's night. The sun acts like a spotlight and shines downward as it moves. The picture below illustrates how the sun moves and also how seasons work on a flat earth.

According to this, and if the earth were flat, then the sun could be seen always, even if it is night, because there would be no possible obstacle for the rays of the sun to travel to the infinite.
Because we cannot see the sun during night, the FE model proposes a solution for this.

According to the FE model, the rays of light from the Sun bend up, so they can't reach our eyes beyond the terminator.

Quoting what the FE wiki says:

Quote
(http://wiki.tfes.org/Optics#Electromagnetic_Accelerator)
Electromagnetic Accelerator
The Electromagnetic Accelerator Theory calls for light to be "bent" upwards as it travels towards the earth. The path of light is a parabolic arc. It is commonly abbreviated to EA.

Day/Night with Electromagnetic Accelerator
When the sun is too far away rays are bent in a parabolic arc before they reach earth, resulting in night time.

The Electromagnetic Accelerator (EA from now on) is compatible with the Night and Day cycles, and it even can give an explanation why the sun can't be seen from a place at night.

But, how do we can calculate the Y position of a photon while traveling?

The FE model kindly gives us a formula to calculate this:

(http://wiki.tfes.org/Electromagnetic_Accelerator)

where X and Y are the position of the photon; c the speed of light in a vacuun, and β is the Bishop constant, that gives the magnitude of the acceleration.

This formula per-se, doesn't make any sense, since in the FE model as depicted in the wiki, doesn't exist gravity as "attraction between two bodies due to their mass", and the formula talks about accelerating a ray of light, but the only way to accelerate the light is to change its relative distance from an object with mass that has gravitational force, but since the FE model the gravity is not an attraction force, then the photons can't accelerate in either direction.

There is also the Davis Model, where it is stated that the Earth do have gravitational force as everyone knows today, and it is an atraction force. In this case, there is gravitational force, so if the photons are being accelerated upwards, it means that there is a gravitational force in the firmament that is able to bend the light. The only way that a mass can bend the light, is if it's gravity is infinite (yes, like in a black hole), so in this case, the existence of the earth itself is a contradiction, because if the light is bent up by infinite gravity, then the source of gravity is close enought to attract everything that has mass and is near the source of light!!!

This, automatically invalidates the EA and everything derived from it. But since I'm not trying to demonstrate that now, lets just simply accept that the photons accelerate because they have a rocket attached to them.

In either case, the Sun illuminates from the Arctic to the Antarctic, which is a distance of ~6300 Km.
This means that a photon that starts travelling from the Sun, parallel to the earth surface, can be seen by a person that is at the same heigh as the sun (4828 Km), and at a maximun distance of ~3150 Km from the Sun.

The meaning of this is that, beyond ~3150 Km from the sun there is no light from the sun at all. Otherwise, we would be able to see the sun during the night.

According to this, when the distance from the Sun to the Moon is more than ~3150 Km, then there won't be light from the Sun reaching the Moon, so the Moon can't reflect any light from the Sun, making it invisible, and this will happen only during the night.

The logical conclussion is that the Moon, in the FE model as depicted in the wiki, can't be seen during the night because the Moon and the Sun are in the same plane of orbit, and the photons from the Sun in such plane are no longer in such plane beyond ~3150 Km.

But in fact, we see the moon during the night, making again the FE model invalid and contradictory.

Conclussion: according to all logical deductions, and using the FE model formulae, we concluded that such model is wrong again since it states that the Moon can't be seen during nightime, while the reality shows us otherwise.

2
##### Flat Earth Theory / Shadow Object and empyrical demonstration against total eclypses.
« on: July 05, 2014, 06:22:02 PM »
Next in my series of empyrical demonstrations about FE failing to explain common events in the nature, let's demonstrate that full eclipses that project a shadow umbra over a region of the earth, are impossible with the FE model.

If you didn't noticed, when the Moon/Shadow Object moves between the Sun and the Earth, it proyects a shadow, but a shadow has different parts, that you can see in the next diagram:

I will demonstrate that eclipses where the whole Solar disc is hidden by the Moon/Shadow Object, are impossible according to FE model, but the real thing is that we can see a lot of eclipses of this type along the history of the earth, as well as predicting when, where and how they will happen.

So, lets get into matter.

First, let me pressent you a simple diagram of how FE illustrates the Earth, Sun and Shadow Object.

In this image, you may note that the Shadow Object (blue small circle) is not in the same plane as the Sun (yellow circle) orbyts around the Arctic.
This is because if the Shadow Object is in the same plane as the Sun, then eclypses won't be possible at all since we will be seeing the Sun from a lower angle that the Shadow Object can cover part of the Sun.

Now, for umbra eclipse (we will call "umbra eclipse" to those eclipses where the whole solar disc is hidden by the Moon/Shadow Object), we need to stay observing from the umbra zone.

Everyone can agree that if we have a source of light and an object obstructing the light (this object is smaller than the source of light) rays falling upon a surface, we can change the umbra, penumbra and antumbra by moving closer and further the blocking object.

Actually, when the blocking object is close to the light source, the umbra length becomes smaller, and when we move farther from the light source the blocking object, the umbra length becomes longer.
You can experiment this in your own house using a light bulb and a iron marble or any other similar opaque object tied to a thin thread.

So, translating this to a Sun-Moon-Earth scale, the Earth is our "wall", the Moon/Shadow Object is the obstructing object and the Sun is the source of light.
To see an eclipse where the Moon/Shadow Object covers the whole Solar disk, we need to stay in the Umbra zone.

Let's then calculate, using the FE model sizes and distances, the length of the Umbra zone, to see if it is possible that the Shadow Object predicted by the FE, allows or not a Umbra Eclypse in the earth surface.

From now on, lets call "Moon" to the Shadow Object, for the sake of abbreviaton.

What we need to calculate is B distance, in the previous image (note that the previous image may not be at a correct scale).
Using some basic school trigonometry, we know that:

B = RMoon/Sin(α)

RMoon (remember, Shadow Object), we know it's value according to FE theory: from 4 to 8 kilometers. Because FE scientists doesn't seem to reach an agreement, lets take the medium size: 6 kilometers.

But we need to calculate the angle "α".

For this, we will use again simple school maths:

α = Arcsin(RSun - RMoon / DSun - DMoon)

Where RSun is the Radius of the Sun (25.74 Km in the FE model), DSun is the distance from the Earth to the Sun and DMoon the distance from the Earth to the Moon (obtaining here the distance from the Sun to the Moon, again basic school Maths).

According to FE model, the distance Sun-Earth is 4828 kilometers.
The distance from the Shadow Object to the Earth or the Sun is unknown, so we need to "invent" a distance. Since we know "the Shadow Object is very close to the Sun", then lets say the Shadow Object is at 4728 Kilometers from the Earth (which is then 100 kilometers from the Sun. Please note that "very close" is a subjective description, but for a 4828 Km distance, it could be agreed that 100 kilometers is very close.

So giving numbers to all of this, we get:

α = Arcsin(25.74 - 6 / 4828 - 4728) = Asin(0.19) = 10.95º

Now, we can calculate the length of the umbra zone for the FE model, given the case that the Shadow Object is 100 kilometers away from the Sun:

B = 6 / Sin(10.95) = 6 / 0.18 = 33 Km

So, we have that the umbra zone is only 33.33 kilometers long.

But according to the FE model, the Sun is at a distance of 4828 kilometers above the earth surface, and given the Shadow Object that is "very" close to the Sun (we supposed 100 kilometers), then we have an umbra zone that goes from 4728 Km to 4694.67 Km above our heads.

So actually, this means that we will have to travel 4694.67 Km up in the sky and firmament to enter the umbra zone, or what is the same, the distance from the Shadow Object to the Sun is so low that the umbra zone projected doesn't even come any close to the Earth Surface.

In either case, this calculation we just made, is actually invalid, since we don't know the distance from the Shadow Object to the Sun. It was stated as "very close", but this is a subjective description. I can say very close when it is 1 milimeter away, and other may say very close for 1 centimeter away.

So what we need to find is a distance A (in the image above) that gives a distance B enought to reach the earth surface.

Getting back to α (alpha), we have:

α = Arcsin(25.74 - 6  / 4828 - Y)

Being X the distance from the Shadow Object to the Earth, which we want to find.

If we take Y as 1000 Kilometers above the Earth Surface, we have:

α = Arcsin(25.74 - 6  / 4828 - 1000) = Asin(0.0051) = 0.29º

And finally: B = 6 / Sin(0.29) = 1185 Km

So having the Shadow Object at a distance of 1000 Km above the Earth Surface, will be slightly enough to allow the surface to enter the umbra zone.

We can conclude then that the Shadow Object must be roughtly 3828 Km away from the Sun, which, for me and most of the humans in the earth is not a "very close" to the sun, if we compare that "very close" with the distance from the Earth to the Sun.

Actually, if we draw an illustration to scale, we have the next (Sun and Shadow Object diameters are not at correct scale respect distances, it would be impossible to draw in a small image like the next):

Which for the Flat Earth model, seems wrong.

Conclussion: following the FE model, and even when there is no a complete source of distances information, the FE models seems to fail when explaining Solar eclipses, since in a FE world won't exist full solar disk eclipses.

3
##### Flat Earth Theory / Sun and empyrical demonstration against FE
« on: July 05, 2014, 11:31:16 AM »
Today I'm going to demonstrate that according to everyday's observation made myself and by every human over the Earth surface, the Sun, Moon and Planets orbyt around the north pole in the flat earth, is wrong, contradictory with the reality.
Due to this, every hypothesys in the FE model derived from (or using as base) the orbyt of such bodies, becomes automatically invalidated, such as the Shadow Object, the Time Zones, the Seasons, the Close Planets Elongation, etc...

So, lets get into matter.

First we will start reviewing what the Flat Earth model teaches us about the Sun orbyt.

According to the most accepted model of the FE, the Sun (and the Moon and Planets) rotates around an axis that is perpendicular to the Arctic. He (the Sun) rotates in a fixed plane that is parallel to the Earth Surfate in their Flat Earth.

The TFES wikia depicts it with the next animation graphic:

According to this, the Sun will be moving in the same movement plane respect to the earth surface along the day and year. The only change is the radius of the orbyt, explaining the seasons this way, but other that this, the Sun's orbyt doesn't have any other change.
What this means is: no matter the time of the day or the day of the year, we will always see the sun at the same height respect to the surface (or respect to any given point in the surface) and respect to the horizon.

But also we can clearly deduct that from a tall mountain, for example the topmost part of the Everest, where there are no buildings or any other obstruction interrupting our horizon view, we will be able to see the Sun when it is midday in the Everest, but also when it is midday in New York.
It is possible that someone can argue that the atmosphere thickness and low transparency, as well as the weather conditions or the atmosphere distortion or other similar effects, doesn't allow us to see the sun even with a normal telescope or our unaided eyes, but even if we couldn't see it, we can observe other sun radiation like X-Rays or any other radiation, which is unaffected by atmospheric distortion or low transparency, so with any appropiate equipment, we will be able to observe the sun from the Everest when it is in New York.

If we take the previous animation and change the perspective, we get the next:

Here, the white thick line represents the supposed flat earth, the red line is the orbyt of the sun around the North Pole movement axis.

If we accept as truth the FE theory about the Sun's orbyt, now it is clear that an observer located in the left of the Earth surface, can see or detect the Sun when along its whole orbyt, without changing his observation possition.

But actually, the FE model states as universal truth this hypothesys without any kind of proof, not even photographs, radiation scanners, radio receivers, not even any kind of mathematical model that explains not only the circular movement of the Sun, but also explains why this movement is round, by which force or mechanism the Sun keeps "tied" to the orbyt instead of being thrown by centrifugal force, or even by which force or mechanism the Sun orbyt radius changes along the year to produce the seasons.

But anyway, if we accept that hypothesys without any proof because "yes", then again, the reality is showing us a Sun's behavior that clearly contradicts what the FE hypothesys says about the Sun orbyt.

Demonstrating this is as easy as tracking the Sun movement respect  the horizon, which anyone can do from his/her home:

Lets check the next image:

Here, we can see how everyday the sun appears from low in the horizon, seems to go up respect the horizon, and then starts descending towards the horizon.
Of course, we don't know if the sun is closer or farther respect us, because we cannot directly see the change in size of the Sun (it can be measured with other methods, but this is not the matter of this topic now). What we see is that the Sun, along the day, seems to be closer or farther from the horizon line, instead of being closer or farther from just us.

So, the Flat Earth model predicts a Sun possition and orbyt that doesn't matches what we see everyday.

Conclussion: due to what I demonstrated before, the Sun's, Moon's and Planet's Orbyts predicted by the FE model are incorrectly defined, besides that, every hypothesys that needs the FE Sun/Moon/Planets orbyts for its explanation or demonstration, is now invalidated too since they start from an invalid hypothesys.

4
##### Flat Earth Theory / Rory Cooper and Celestial movement
« on: July 04, 2014, 12:08:44 PM »
Found in this thread, pizaaplanet points us to a video made by Rory Cooper that explains about how the stars move (or seem to move) in the firmament.

The video is the next one:

Most of the video is wrong because it has some terrible errors that doesn't match what we observe everyday.

First, the video assumes that the earth is flat, which is an assertion that cannot be demonstrated, making the rest of the video invalid (you cannot state something as truth if you start from a hypothesys that you cannot demonstrate).

But even if we don't have this into account, Rory Cooper commits a serie of errors that made the video invalid no matter what you try to accept, since those errors aren't, by far, what the reality shows us.

First, at 0:44 he asumes that the earth is standing still, which is, again, an assertion that isn't demonstrated, making again the whole video invalid.

Next, in 0:58 he shows us several hypothesys why the stars move or seem to move. The only possibilities are that the observer is over a spinning earth or the stars are spinning. But since he stated before that the earth is standing still, he is contradicting himself by saying now that the earth can spin.

But now, at 1:16 negates something that is actually happing, and can be evidenced by anyone: he states that the stars must be spinning around the earth's rotation axis, which is what we can see if we travel to the north pole.

If we travel to the north pole and record the stars movement above our heads, we will see this:

Which is perfectly correct and doesn't represent any contradiction between our round earth theories and what we can observe.

But now, if we travel closer to the equator, but still in the northern hemisphere, the thing changes.
If we look at Polaris from our new location, we well see the next:

And actually, this is truth since if you travel from, lets say, Sahara Desert to North Sweeden, you will clearly see that Polaris (and every star) seem to be higher (or lower) above the horizon.

I was not in Sahara Desert, nor in Sweeden, but I was in Morocco and in North of UK, and the diference between the possition of stars in both places are clearly noticeable.

Now you may guess how this can happen.
The most plausible answer is this, which, in fact, matches all observations:

Let me explain what this diagram means:

We are first observing the night sky in the north pole. It doesn't mind if the earth is standing still or rotating, or are the stars who move.
If we look to polaris, we will see it just above our head because we are just in the movement axis, and everything star will seem to rotate around Polaris (actually not exactly Polaris).

Now, lets travel to closer to the Equator but still in Northern hemisphere (the red astronomer in the diagram). To see Polaris, we will need to face towards the movement axis, which is in the North Pole. Now, we can observe that Polaris appears way lower in the horizon.
We will see the stars spinning anyway, so I can't see any contradiction between a Round Earth and how stars seem to move in the sky.

Now, at 2:15 Rory Cooper says that some people stated that it is possible to see stars from the north and south hemisphere from the equator. The only possible way is the earth being a plane. But in this case, I will be able to see south stars from anywhere above the equator, which is not possible.
But, if you travel to the equator in a round earth, thanks to different earth movements, it is possible that the most southern stars and northern stars are visible during different seasons, of course, not during the same day.

At 2:47, Roory Cooper starts to talk about the universe as a time piece to measure the months, years, etc.., but you cannot universally measure something that is relative.

Also without having into account that the concept of time is something abstract that was created by the humans to measure and take account of events that happen regularly. Talking about time is completely and utterly pointless.

I can even say that time is just a concept like "love" or "hate", that doesn't have any impact in the reality, nor it is related to any universal physical fact or object, not even can be measured by any means.

At 3:09 Rory Cooper states that the moon is too far to see any kind of detail, which is an statement without demonstration or any other kind supporting data.

So, taking all those true facts I exposed before, Rory Cooper's video is wrong, away from the reality and terribly confusing.

5
##### Flat Earth Theory / Moon and empyrical demonstration against FE.
« on: July 03, 2014, 03:16:10 PM »
Distance and size to the Moon in your model is wrong. I can prove it and give true information (e.g. anyone can check the validity, and I already did using my own equipment, which I describe below)

Using any stargazer telescope (I own several telescopes of several type), a digital camera and a computer you can easily measure the size of Moon's features like craters, mare, rims, etc...

Let me post a Moon image that I took some some months ago:

In this image, you can see the crater Copernicus, which is a crater that is near the Moon's equator (actually roughtly 150km above the equator). The estimated size of the crater is 93kilometers diameter (is a round crater). There are also other craters, mons and features around, but due to the low budget camera a telescope used, there are plenty of them that can't be seen because they are too small to be seem from this distance with the equipment i describe next. The smaller features that I can see with this telescope and camera is roughtly 4 kilometers diameter. Something smaller will appear only be visible as a very small fuzzy shadow (or light reflection).

You may say that someone made us believe that Copernicus is 93kilometers diameter, or that I'm saying fake information, so let's study the matter to understand if that is true or false:

The image was taken with a galilean telescope (uses lenses) with a focal distance of 1000 milimmeters and an aperture of 100mm.

The image was taken using a digital camera, but let's start supposing we are first looking through that telescope with an eyepiece with a focal lenth of 10 milimeters. (If you don't know what all this means, then please stop reading, since you won't understand a lot of concepts beyond this line)

Now, according to basic optics, using such eyepiece in such telescope, we will be seing objects as if they were 100 times closer to us.
So, focusing the telescope to a tree that is 100 meters away, will look like another tree that is 1 meter away from us.
You can't negate this, since you can check it yourself (I already did, and it is true).

Now, if we focus the telescope to the Moon, if it is 394.400 kilometers away, we will see the moon as if it were at a distance of 3944 kilometers instead of 394.400.

This can't tell us if the Moon if big or small, but just let us measure distances using equivalences when using a digital camera.

Now, let's switch to the camera I used to get the image above. It was a Luna-QHY 5L-II CMOS camera.
This camera has a pixel size of 3.75 microns (squared pixels), a 1/3 inc sensor (8.43 mm squared sensor) and without any binning (binning 1x1 actually)

Now using basic informatics and optics formulae, we got a magnification of x83.58 , and an apparent FOV of 29x29'/pixel.

Now, according to the FE model, the distance Earth-Moon is 15 kilometers and has a 600 meters diameter.
So, if I take a photo with that telescope and that camera, the moon will look as if it were 0.15 kilometers away (150 meters away). As for comparison purposes, if I look at a house that is 150 meters away, I can see every kind of details that are even 1 meter diameter, lets say, a kid playing around. Please, take this into account because it is very important, even when it may seem otherwise.

Now according to RE model, the distance Earth-Moon is 394.400 kilometers and 3475 diameter.
Using the telescope and camera above, it will look like 3944 kilometers away. Now, if I look at a house that is 3944 kilometers away, i can't really see it because it is too small and the distance too big.

But now lets take into account this fact: using that telescope, it is impossible to see craters or features that are roughtly 4 km diameter or less because of the distance and the telescope resolution. This is an evident fact that you can check it's validity if you know a little bit about optics and geometry. But again, doesn't seem to work in your FE model, so in your FE model with a moon at a distance of 15 kilometers, I could see even the american flag or the footsteps of the astronauts, but truth is that I cannot see anything that is smaller than (roughtly) 4 Kilometers.

But when I use any of my high-end telescopes, I can clearly see details from the Moon surface that are even 1 or 2 kilometers size, so not seeing them with the low-budget telescope is not due to the inexistance of such craters features, but due to the lack of telescope quality, while in your FE model, any telescope (even cheap chinese plastic ones) will let us to see moon details that are as small as 1 or 2 meters diameter in size, but this actually doesn't happen.

Conclussion: according to everything exposed and demonstrated previously, the distance from the Earth to the Moon is wrong in your model, as well as its size, since it doesn't match my observation or the observation that anyone can do at their home with the appropriate equipment.

6
##### Flat Earth Theory / This can't happen if earth is flat
« on: July 03, 2014, 12:16:30 AM »
Hi.

Not sure if this was posted somewhere before, but just want to ask how can you explain for your flat earth model:

1- that I can cross the south pole without falling to the void. Lets say, I walk towards the south pole from the Pacific Ocean, I cross the pole and I reach Atlantic ocean. Your flat earth model is against the own nature of these evident facts at first sight.

2- If the Sun is as small as you state (I cannot remember now the size in your model), its gravity will be so small that gravitational lensing due to the Sun's gravity would be impossible to observ, but in fact, it exists and it can be empirically demonstrated, and even ovsersed with an appropriate telescope during an eclypse. How your flat earth model can demonstrate or explain this?

3- Astronomical objects that "orbyt" the solar system from a far distance, such as comets, have a perion in which they can be seen from the earth with just an optical telescope. During their transit near earth, they are visible only in some parts of the earth, for example, they can be seen the first 50 days in the north hemisphere, and the other 50 days in the south (supposing their transit is 100 days). How can you explain or demonstrate for your flat earth model that such objects from that far distance are hidden from certain regions in the earth?

4- In your model there is a contradiction between sun's size and energy irradiated. How can you explain that such a small sun can irradiate energy to reach the zones that are at down or dusk, without destroying the zones that are at mid-day?

5- when I observe certain planets with any of my telescopes, there are days when they seem bigger during a certain period of the year. This period changes each year (maybe in 2014 is during september but in 2015 is during April).
This is because such planet is closer to the earth due to their eliptycal orbyts around the sun (in a round earth heliocentric model), and this can be seen from everywhere in the Earth during that period, no matter the season or any other factors. In a flat earth model, this cannot happen, since a point in the space can't be equidistant to every point in a flat surface, unless your surface is a concave object, and in this case, the distance to Jupiter or any other planet will greatly and noticeablely vary depending on where in the flat earth surface I am. But according to any common observation, the distance to Jupiter at naked-eye or even using a telescope, looks the same even if I am in the north pole or in the equator.