Issues with sun location on North-pole-centred Flat Earth model presented in FAQ

I opened up Stellarium and found the azimuth/declination of the sun and different times of the day on September 21 2010.  Using this data, I calculated the distances and altitudes the sun would have to be from Ottawa, Canada if it was following a path over the equator on Flat Earth.

Stellarium is a program that shows you exactly where in the sky any planet, or major moon or asteroid would be at any time from any place in the world, or on any other planet, major moon or asteroid down to the last 2 decimal places of an arc second.  Feel free to test the accuracy of it yourself, as it's a free download.


As you can see, the math shows that the altitude of the sun varies from 0 km at sunset or sunrise, to 5130km at 1:00 PM.  This is contrary to the idea that the sun maintains a constant altitude.

Another problem I notice is that throughout the 12 hour period, the sun covers less than half of it's total path, which would be impossible if it were supposed to travel at a constant speed making 1 revolution every 24 hours.

Also, if I were to attempt this diagram showing the sun's position as viewed from south of the equator, even more problems would arise.  Some of the vectors may not even coincide with the sun's alleged path.

No doubt they will try to somehow explain this by invoking "bendy light", or FE's answer to just about everything it doesn't have a working explanation for.

Parsifal's EA is out, as this problem would require a horizontal bending of light as well as an upward curve.   The observation, however, is consistent with Round Earth Theory.

Parsifal's EA is out, as this problem would require a horizontal bending of light as well as an upward curve.   The observation, however, is consistent with Round Earth Theory.

As if it was ever really in.

I opened up Stellarium and found the azimuth/declination of the sun and different times of the day on September 21 2010 (fall equinox).  Using this data, I calculated the distances and altitudes the sun would have to be from Ottawa, Canada if it was following a path over the equator on Flat Earth.

Stellarium is a program that shows you exactly where in the sky any planet, or major moon or asteroid would be at any time from any place in the world, or on any other planet, major moon or asteroid down to the last 2 decimal places of an arc second.  Feel free to test the accuracy of it yourself, as it's a free download.


As you can see, the math shows that the altitude of the sun varies from 0 km at sunset or sunrise, to 5130km at 1:00 PM.  This is contrary to the idea that the sun maintains a constant altitude.

Another problem I notice is that throughout the 12 hour period, the sun covers less than half of it's total path, which would be impossible if it were supposed to travel at a constant speed making 1 revolution every 24 hours.

Also, if I were to attempt this diagram showing the sun's position as viewed from south of the equator, even more problems would arise.  Some of the vectors may not even coincide with the sun's alleged path.

These problems come up when you assume that the earth is a globe.

It's not.

These problems come up when you assume that the earth is a globe.

It's not.
[/quote]

Trolllllll alert. All I got from this post was "This evidence firmly supports round Earth, but I don't believe it so it aint true."

Quote from: Nolhekh on October 21, 2010, 05:44:42 PM

I opened up Stellarium and found the azimuth/declination of the sun and different times of the day on September 21 2010 (fall equinox).  Using this data, I calculated the distances and altitudes the sun would have to be from Ottawa, Canada if it was following a path over the equator on Flat Earth.

Stellarium is a program that shows you exactly where in the sky any planet, or major moon or asteroid would be at any time from any place in the world, or on any other planet, major moon or asteroid down to the last 2 decimal places of an arc second.  Feel free to test the accuracy of it yourself, as it's a free download.


As you can see, the math shows that the altitude of the sun varies from 0 km at sunset or sunrise, to 5130km at 1:00 PM.  This is contrary to the idea that the sun maintains a constant altitude.

Another problem I notice is that throughout the 12 hour period, the sun covers less than half of it's total path, which would be impossible if it were supposed to travel at a constant speed making 1 revolution every 24 hours.

Also, if I were to attempt this diagram showing the sun's position as viewed from south of the equator, even more problems would arise.  Some of the vectors may not even coincide with the sun's alleged path.

These problems come up when you assume that the earth is a globe.

It's not.

Then explain how its possible on a flat earth.

These problems come up when you assume that the earth is a globe.

It's not.

We shall see about that...

Quote from: Tom Bishop on October 21, 2010, 08:23:25 PM

These problems come up when you assume that the earth is a globe.

It's not.

We shall see about that...

So wait...

Evidence is given that supports a round earth model. Given that I'm sure the program the OP used is consistently accurate, how is assuming the earth is a globe a problem?

A good theory allows you to make good predictions. If you can accurately track the positions of the sun, moon, planets, etc based on a round earth model, what does this say about the FE model? 

 It was automated response, in reality Tom didn't understood what the problem was.

These problems come up when you assume that the earth is a globe.

It's not.

Wow, Tom.  That is a very compelling and well thought out response.  I am convinced.

Typical Tom Bishop response.

Quote from: Tom Bishop on October 21, 2010, 08:23:25 PM

These problems come up when you assume that the earth is a globe.

It's not.

We shall see about that...

Round Earth Theory tells that the sun maintains an average altitude above the equator of approximately 150 million km, so there is no problem with the sun having an altitude of 0km

As for the sun being able to complete it's path, there is no problem.  The sun itself, according to RET, does not have a path, but it's apparent motion is due to the earth rotating.  The sun's apparent 180 degree motion across the sky in 12 hours means the Earth rotates 180 degrees in 12 hours, giving it 12 more hours in a 24 hour day to rotate the next 180 degrees the rest of its 360 degree rotation, so there is no speed problem.

Another thing, notice that when the sun reaches it's southernmost position at 1300, it has a declination of 45 degrees, which matches the 45 degree latitude of the location.  This shows a reasonably undeniable consistency with round earth theory.

These problems come up when you assume that the earth is a globe.

It's not.

Tom, in the over 2 years that I've been here, there have been several attempts to organize an experiment where FES members would observe the sunrise and sunset on the day of the equinox.  Have you ever participated in one of these experiments?  It seems like an easy way to determine which model (FE or RE) is better supported by real world observations.

Round Earth Theory tells that the sun maintains an average altitude above the equator of approximately 150 million km, so there is no problem with the sun having an altitude of 0km

I realise that this may require some explanation:
When the sun reaches 0 degrees declination on round earth, it doesn't mean it's 0km off the ground like it would for flat earth.  This is because the ground curves away in the round earth model, and where the ground faces the sun precisely, the 150 million km altitude is still accurate.

New figures regarding the size of the sun have been calculated, and show that following the equatorial path in this model, the sun would have to change size throughout the day to maintain a relatively constant angular size while following a path that changes it's distance from the observer.  The sun does change angular size over the course of one year ranging from 31.6 arc minutes on July 3 to 32.7 arc minutes on January 3. The RE explanation for this is that because of Earth's elliptical orbit, the sun is further away during the months surrounding July, and closer during the Months surrounding January. Since I'm not yet sure how to calculate where in the orbit something is at any given time, I can't say exactly what the sun's angular size would be on September 21, so I calculated the range of sizes it could be at a given hour.

Time      Size range in km     Size Range in Miles
0653      80.59 -   83.4   50.08-   51.82
0700      79.74 -   82.52   49.55 -   51.28
0800      73.13 -   75.68   45.44 -   47.02
0900      69.07 -   71.47   42.92 -   44.41
1000      66.93 -   69.26   41.59 -   43.04
1100      66.01 -   68.3   41.01 -   42.44
1200      65.81 -   68.11   40.9 -   42.32
1300      66.21 -   68.51   41.14 -   42.57
1400      66.2 -   68.51   41.14 -   42.57
1500      66.44 -   68.75   41.28 -   42.72
1600      67.38 -   69.73   41.87 -   43.33
1700      69.71 -   72.14   43.32 -   44.83
1800      73.96 -   76.54   45.96 -   47.56
1859      81.34 -   84.17   50.54 -   52.3

I've created an animation of the Round Earth, rotating as expected in RET.  Using the Azimuth/Declination data shown in the OP, I added vectors to Ottawa's coordinates showing the direction of the sun.  The sun's vectors are Red (Sunrise and sunset) and Yellow.  You can see, as the Earth rotates, the solar vectors stay pointed in the same direction, just as expected in RET.  This demonstration was 100% calculated using data given by RET in conjunction with real life observable celestial movements, and shows that the two fit together perfectly.

Would a Flat Earth Theorist please provide a Flat Earth Model that is mathematically consistent with real life celestial observations as Round Earth Theory has, or suggest a specific part of "Earth Not a Globe" or other book or reference which provides such a model?



P.S.  to Round Earthers, please do not post that they will never be able to do this.  It will only result in trolling and low-content posts which will in no way contribute to answering the question I have posed.

A real life observation from the medical point of view is that that type of spinning will cause nausea. I defer to the Specialist FE Theorists to respond to your specific needs.   

A real life observation from the medical point of view is that that type of spinning will cause nausea. I defer to the Specialist FE Theorists to respond to your specific needs.    

1 revolution every 24 hours causes nausea?  This rate of rotation is what my animation represents.  This would result in 0.022382 m/s² 44.75^o down from North acceleration on my body.  That's only 0.00243 g's.

A real life observation from the medical point of view is that that type of spinning will cause nausea. I defer to the Specialist FE Theorists to respond to your specific needs.   

If that type of spinning was a problem, then revolving restaurants would have some pretty significant product return issues to deal with.

A real life observation from the medical point of view is that that type of spinning will cause nausea. I defer to the Specialist FE Theorists to respond to your specific needs.   

We don't perceive the spin.

http://curious.astro.cornell.edu/question.php?number=665

Quote from: gotham on October 27, 2010, 06:13:56 PM

A real life observation from the medical point of view is that that type of spinning will cause nausea. I defer to the Specialist FE Theorists to respond to your specific needs.   

We don't perceive the spin.

http://curious.astro.cornell.edu/question.php?number=665

once you reach the north or south pole it disappears completely because the Earth is not spinning there.

This statement is incorrect.  The earth spins at the poles.  It would be better to say that the surface does not follow a circular path, and thus no outward force is experienced. However, there is centrifugal force on the sides of whatever object would be sitting right on the pole.  A lot less force, but still there.

Quote from: gotham on October 27, 2010, 06:13:56 PM

A real life observation from the medical point of view is that that type of spinning will cause nausea. I defer to the Specialist FE Theorists to respond to your specific needs.    

We don't perceive the spin.

http://curious.astro.cornell.edu/question.php?number=665

Editing my post because I didn't lurk moar.  I see this is all covered in a recent thread about the controversial earth or sky spinning issue.

nice job

and you can add places that are aligned to the sun

such as stonehenge, ceratin pyramids, solar calandars, ect