They told us the sun is 93 million miles away, but we don’t see evidence for that. It is what it looks like: a small, local, mobile light. As he travels his circuit across the sky, different regions experience daylight. He lights up the clouds around him, so you can tell he’s close!

THE SUN IS:
• Small and local • Moving all the time • Marking days, seasons and years

THE SUN IS NOT: • A giant gas ball • The center of our system • Ninety-three million miles away

I will try to explain why the sun cannot be so close as FE’s says.
All the values are real proving that it’s impossible the supposed position of the sun in a FE.
For this calculations the following elements were used:
1. Stellarium – Software that tells you the position of celestial bodies with accuracy. (http://stellarium.org)
2. Google Maps – To measure distances. (https://www.google.com/maps)
3. GPS Coordinates – To see the coordinates of the points to measure (https://www.gps-coordinates.net/).
4. Trigonometry.
First, we need to calculate the sun altitude:
We need two points on earth and the sun elevation angles in this two points: (this calculations were made on March 20, 2018 at 16:21:09 GMT)
The coordinates of the first point (a) are: 0°0’0”S, 78°26’19”W and the angle to the sun is: 89°58’39” ≈ 90° (This variation does not significantly affect the experiment.)
The coordinates of the second point (b) are: 48°37’55”N, 78°26’19”W and the angle is: 41°23’56” ≈ 41°.
The distance between point ‘a’ and ‘b’ is: 5407.62km.
So, we have a right triangle: segment ab = 5407.62, and angle θ = 41°.
To calculate the sun altitude (c):
c = tan(41) * 5407.62
c = 0.869 * 5407.62
c ≈ 4,700 km (2,920 miles, almost the 3,000 that I’ve heard from people who know about this)
4,700 km must be the sun altitude in a flat earth.

Now, we need to know the elevation angle of the sun when it is night. This angle must be greater than zero (because the sun is always up) and lower than zero (because, otherwise, it should be visible… wait, we have a contradiction here… with this, the conclusión is that the sun cannot be at that altitude, the calculations don’t lie… well, let’s ignore that and proceed).

For this, we need to know the distance between any point in the equator and its opposite, i.e, two points that delimit the equator diameter.
We have a small problem here:
In the globe model, the equator is 40,000 km long, so, the diameter has to be: 12,732 km approximately. The problem is that, the distance between two points that are opposite in the equator is not the diameter, but, half circumference because the spherical shape of earth, i,e, the real distance is approximately 20,000 km. So, we have two different distances here (this is another proof of the wrong flat model). Let’s use the bigger one to try to help flat earthers (greater this value, lower the elevation angle of the sun at night)

So, we have another triangle here:
Point ‘a’ is some place near Quito, Ecuador, point ‘b’ some place near Indonesia and point c is the sun.
What is the elevation angle of the sun from point ‘a’ when the sun is above point ‘b’? i.e. when it is night.
It is a right triangle: segment ab = 20,000 and segment bc = 4,700.
θ = atan(4700 / 20000)
θ ≈ 13°13’28”

So, you need to look towards north and turn your head 13 degrees up to see the sun at night.
Do you know how much thirteen degrees are?
You can download Stellarium (is a totally free software), put your location in the software, turn on the azimutal grid to see the angles lines (is a button in the lower menu with a concentric circles image), and move the time until the sun gets the 13° line (the sun touch this line in two moments of the day: at sunset and sunrise) and see what time is at that moment, then, you need to wait until that moment of the day and look at the sun (the real one, not the sun in stellarium) and that is how you would see the sun when it is night (towards north).

I hope that you people can make this experiment of see the sun at 13 degrees with stellarium help.

As I said before, all the values are real, if anyone thinks that there are some wrong value, let me know to correct the calculations.

I will try to explain why the sun cannot be so close as FE’s says.

All the values are real proving that it’s impossible the supposed position of the sun in a FE.

For this calculations the following elements were used:

1. Stellarium – Software that tells you the position of celestial bodies with accuracy. (http://stellarium.org)

2. Google Maps – To measure distances. (https://www.google.com/maps)

3. GPS Coordinates – To see the coordinates of the points to measure (https://www.gps-coordinates.net/).

4. Trigonometry.

First, we need to calculate the sun altitude:

We need two points on earth and the sun elevation angles in this two points: (this calculations were made on March 20, 2018 at 16:21:09 GMT)

The coordinates of the first point (a) are: 0°0’0”S, 78°26’19”W and the angle to the sun is: 89°58’39” ≈ 90° (This variation does not significantly affect the experiment.)

The coordinates of the second point (b) are: 48°37’55”N, 78°26’19”W and the angle is: 41°23’56” ≈ 41°.

The distance between point ‘a’ and ‘b’ is: 5407.62km.

So, we have a right triangle: segment ab = 5407.62, and angle θ = 41°.

To calculate the sun altitude (c):

c = tan(41) * 5407.62

c = 0.869 * 5407.62

c ≈ 4,700 km (2,920 miles, almost the 3,000 that I’ve heard from people who know about this)

4,700 km must be the sun altitude in a flat earth.

Now, we need to know the elevation angle of the sun when it is night. This angle must be greater than zero (because the sun is always up) and lower than zero (because, otherwise, it should be visible… wait, we have a contradiction here… with this, the conclusión is that the sun cannot be at that altitude, the calculations don’t lie… well, let’s ignore that and proceed).

For this, we need to know the distance between any point in the equator and its opposite, i.e, two points that delimit the equator diameter.

We have a small problem here:

In the globe model, the equator is 40,000 km long, so, the diameter has to be: 12,732 km approximately. The problem is that, the distance between two points that are opposite in the equator is not the diameter, but, half circumference because the spherical shape of earth, i,e, the real distance is approximately 20,000 km. So, we have two different distances here (this is another proof of the wrong flat model). Let’s use the bigger one to try to help flat earthers (greater this value, lower the elevation angle of the sun at night)

So, we have another triangle here:

Point ‘a’ is some place near Quito, Ecuador, point ‘b’ some place near Indonesia and point c is the sun.

What is the elevation angle of the sun from point ‘a’ when the sun is above point ‘b’? i.e. when it is night.

It is a right triangle: segment ab = 20,000 and segment bc = 4,700.

θ = atan(4700 / 20000)

θ ≈ 13°13’28”

So, you need to look towards north and turn your head 13 degrees up to see the sun at night.

Do you know how much thirteen degrees are?

You can download Stellarium (is a totally free software), put your location in the software, turn on the azimutal grid to see the angles lines (is a button in the lower menu with a concentric circles image), and move the time until the sun gets the 13° line (the sun touch this line in two moments of the day: at sunset and sunrise) and see what time is at that moment, then, you need to wait until that moment of the day and look at the sun (the real one, not the sun in stellarium) and that is how you would see the sun when it is night (towards north).

I hope that you people can make this experiment of see the sun at 13 degrees with stellarium help.

As I said before, all the values are real, if anyone thinks that there are some wrong value, let me know to correct the calculations.

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