21st August solar eclipse on the flat Earth

Today I want to check if the flat earth model  I have till now presented is ok. Especially in the case of a solar eclipse.

The model to keep in mind is that of a flat earth pond on which you should visualize the trajectory of the sun. This is  a cone having the following geometrical data:

SUN Cancer tropic Capricorn tropic
Radius of the orbit 6660 13320
Height of the orbit 6660 3330

The moon trajectory is pretty similar. It runs a cone that is immediately near to the cone of the sun:

MOON Cancer tropic Capricorn tropic
Radius of the orbit 6642 13284
Height of the orbit 6642 3321

These two cones intersect regularly due to the libration movement of the moon. We are going to consider the two average trajectories and present their corresponding data tables. Results should be a good approximation of the behavior of the sun and the moon orbiting the flat Earth.

First I’ll start discussing the phenomenon of the solar eclipse. According to the mainstream science, this sort of eclipse can easily be understood. It occurs due to the passage of the moon in front of the sun. In that case, the moon fully or partially blocks or occults the sun. Below you have a picture of a globular, mainstream model of the solar eclipse.

On a flat earth model, the solar eclipse develops the same way: it’s a passage of the moon in front of the sun. The major differences with mainstream astronomy are that, during the eclipse, the sun and the moon are very near one to the other, they are quite small (they have the same dimension) and are quite near to the Earth. Very likely the sun and the moon don’t have a  mass. You can also read  my post articles about gravity.

Since gravity doesn’t exist in the terms Newton postulates, you can easily conclude that sun and moon, which are pretty near to the Earth, don’t have masses  and  are optical phenomena.

Solar eclipse over the flat Earth

Here below I report an image of our flat earth model with the cone trajectory of the sun.

Now , we have to consider the particular sun eclipse that will be visible on 21 August 2017. This eclipse will be complete on the United States and partial on Canada and Mexico as you can notice from the picture below.

This  eclipse will permit to check the validity of this astronomical model. I will make use of Sunsurveyor . So I will check the reciprocal position  the sun and  the moon will stay in many places of the Earth during the course of the eclipse. Moreover, I will verify if the position of the sun and the moon I have postulated inside this model is coherent with the reality.

The Eclipse as seen from Casper

The first place to start is Casper in the USA. People there will experience a total eclipse at about 12 o’clock, local time.

Let’s check the situation at Casper on 21 August 2017. An observer there will have to watch southward to see the eclipse at 12 o’clock. Sunsurveyor is already foreseeing this position for the sun and the moon. (I have oriented the compass heading south).

If you want to apply the coordinates of the sun and the moon to the flat Earth model simply you have to consider that they are very near and the moon is in front of the sun. The coordinates you can see in the image perfectly work with our model:

156,7° represent the azimuth, i.e. the angle measured from north direction.

58,3° represents the altitude angle.

These two coordinates individuate a star on the firmament. Interesting enough is the fact that you don’t need to know the distance to find the star in the firmament. This is due to the fact that the astronomers imagine all the outer space as projected on an imaginary sphere. Isn’t it the firmament?

 

Based on this model you’ll find  this situation:

The eclipse as seen from Calgary

 

A second place to consider is Calgary in Canada.  To observe the eclipse from Calgary at 12 o’clock, you ‘ll have to watch heading south.

The situation according to Sun Surveyor will be this.

As you can notice the moon is in front of the sun. In our model the situation will simply be this:

The moon is in front of the sun, but not completely. This can’t be appreciated when considering only sun surveyor’s data. We can assert however that with our model it is quite easy to describe the reality.

The eclipse as seen from Quito

Quito in Ecuador is on the equator. On 21 August the sun will be a little northward.

The situation,  according to Sun Surveyor,  will be this:

Notwithstanding the fact that you are quite distant from the zone of the full visibility, even at Quito, the moon will be seen in front of the sun and the eclipse will be partially observed.

Planets coordinates

What about planets? where will they stay during the eclipse? Is this coordinate system used to individuate planets on the firmament as well? Let’s see.

In order to obtain the coordinates of planets, you have to download another application. I’m doing with “Mappa Stellare”, an Italian version of an English application made by Escapist Games that you can easily find on Google play.

I set the position on Casper to compare data I have found with Sunsurveyor. SoI program the date for the 21 August 2017. The time is set according to the Italian timing. For this reason, only, I have to indicate 18 o’clock to obtain 12 o’clock in the United States. This is the situation you will find:

As you can notice, Mars and Mercury will stay at the two sides of the sun and the moon as two best men at a wedding. The moon will be obviously on the ecliptic, that is the sun trajectory. We remember that the moon doesn’t stay on the ecliptic but keeps moving on her cone in 27 days. The entity appearing near Mercury is the constellation of Sextant.

Let’s  now check the coordinates that are given.

As you can verify yourself, the flat earth system coordinates of the moon are very different from those that you will obtain from Sunsurveyor. Why? This kind of software is calculated for a globe geometry so you ought to develop the needed mental flexibility and be adaptable. Granted, the mismatch between the two software shows a remarkable lack of precision.

Anyway, let’s continue.

Mars and Mercury as well are perfectly pinpointed on the firmament by their coordinates. Our model of a semispherical firmament keeps on working.

 

While reading this brief analysis you could have appreciated how far this flat earth astronomical model works in relation with a solar eclipse. What could be said about lunar eclipses? For sure it will be a bit more difficult to explain. To introduce the subject for a future post I’ll leave you, my reader, with an intriguing image. Bye.

6 comments

  1. Are you tired of arguing with Ball-Earthers and getting nowhere? Are you unable to get them to see through the truth’s protective layers? The Flat Earth is a hard sell because the Ball-Earthers have never considered an alternative and all of the propaganda supports what they already believe. They have a vested interest in maintaining their own mental stability so it’s not easy getting them to willingly take on the cognitive dissonance required to wrap their mind around the biggest lie ever told.

  2. Good morning. I’ve read, for example from Eric Dubay, that does exist a black moon that hide the moon light whn in perfect alignment with the sun. From my studies I’m more confident with some electromagnetic phenomenon, for example a sort of magnetic shield that borns when the moon is in the opposition with the sun

  3. So either an undetectable “black moon” or a magnetic shield. OK.
    In the flat earth model what size is the (normal) moon and how fast is it
    travelling and what is the circumference of its orbit?

  4. Hi, I’ve seen your comment only now, sorry. Since the moon covers a 0.5 dgrees angle in the sky and it is at 6-7 thousends km far from us the diameter should be about 65 chilometers. Its orbit is very variable from the tropic of Cancer to that of Capricorn.The radius of tropics is 6660km and 13320km the other

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