Offline Flatout

  • *
  • Posts: 238
    • View Profile
Re: The Sun
« Reply #40 on: March 02, 2017, 08:19:17 PM »
Have you done the calculation yourself to see if that statement is true? 

During equinox on the spherical earth, the sun is directly overheard (90 degrees) at the equator and right on the horizon (0 degrees)  at the north pole.  On the flat model,  with the sun 3000 miles above the earth,  the sun would be 25 degrees above the horizon at the poles during equinox. 

They are not the same.

Lastly, a sextant uses line of sight.  I though you recently stated that due to the complexities of the atmosphere we can use line of sight.  If we can't use line of sight how did TFES calculate the height of the sun?

Re: The Sun
« Reply #41 on: March 02, 2017, 08:47:26 PM »
Have you done the calculation yourself to see if that statement is true? 

During equinox on the spherical earth the sun is directly overheard (90 degrees)  on the equator  and right on the horizon (0 degrees)  at the north pole.  On the flat model with the sun at 3000 miles above the earth the sun would be 25 degrees above the horizon during equinox. 

They are not the same.

Did you even watch the video? It describes how the same principle archemedis used to try to find the distance and diameter of the sun is used to support the flat earth size of the earth.



Here is one about Actual and Apparent position of the sun, if you're curious how reality and perception can differ.

Offline Flatout

  • *
  • Posts: 238
    • View Profile
Re: The Sun
« Reply #42 on: March 03, 2017, 12:48:21 AM »
Yes I watched the video start to finish.  My point is that the statement in the video is incorrect.  Do the math.  It doesn't work.

Your second video is a great explanation about why we can sometimes briefly see a Lunar eclipse and a sun rise at the same time. 
« Last Edit: March 03, 2017, 03:35:19 AM by Flatout »

Offline Flatout

  • *
  • Posts: 238
    • View Profile
Re: The Sun
« Reply #43 on: March 03, 2017, 03:06:43 AM »
Doing the calculations at equinox is easy.  You can even use the wiki to derive your data for the triangulation.  Triangulation is required if you're using shadows or a sextant to calculate height.

Quote
To locate your latitude on the flat earth, it's important to know the following fact: The degrees of the earth's latitude are based upon the angle of the sun in the sky at noon equinox.

That's why 0° N/S sits on the equator where the sun is directly overhead, and why 90° N/S sits at the poles where the sun is at a right angle to the observer. At 45 North or South from the equator, the sun will sit at an angle 45° in the sky. The angle of the sun past zenith is our latitude.

Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.

10*N Latitude, Distance from Equator=695 miles, Suns elevation 80*, Suns Height=3941 Miles
30*N Latitude, Distance from Equator=2085 miles, Suns elevation 60*, Suns Height=3611 Miles
45*N, Suns Height=3127 Miles
60*N, Suns Height=2047 Miles
80*N, Suns Height=980 Miles

So how was the suns height discerned to be 3000 miles?
« Last Edit: March 03, 2017, 03:31:16 AM by Flatout »

Re: The Sun
« Reply #44 on: March 03, 2017, 04:38:48 PM »
Yes I watched the video start to finish.  My point is that the statement in the video is incorrect.  Do the math.  It doesn't work.

Which statement in the video?

Offline Flatout

  • *
  • Posts: 238
    • View Profile
Re: The Sun
« Reply #45 on: March 03, 2017, 10:08:30 PM »
Yes I watched the video start to finish.  My point is that the statement in the video is incorrect.  Do the math.  It doesn't work.

Which statement in the video?

The statement "that the calculations would work out the same on a flat plane with a close sun as on a round earth with a far sun".  Your claim and the claim in the video are not correct....at least in respect to the suns height being 3,000 miles.  If you repeated Eratosthenes' experiment it would yield a sun's height of 4,000 miles. Secondly, your claim that a sextant could calculate the suns height to be 3,000 miles is incorrect.  It reality sextants would yield very different values depending on your latitude during solar noon on equinox (the time that Eratosthenes realized the possibility of his experiment).  In fact, the only latitude that a sextant would yield a value of 3,000 miles is 43* north or south latitude.
« Last Edit: March 03, 2017, 10:16:43 PM by Flatout »