1

Flat Earth Theory / Re: Sunrise and Sunset II

« on: August 10, 2017, 03:58:31 PM »

So, back to my initial question : where am I wrong with my computation with expected distances to get a sun position at midnight 15 degrees above the horizon?

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Flat Earth Theory / Re: Sunrise and Sunset II

« on: August 10, 2017, 02:51:41 PM »

Ok, but if the sun rise/set from/to a specific location, this is not compatible with this animation : http://wiki.tfes.org/File:SunAnimation.gif, and if the sun disapear below the earth surface, all the countries should be in the night at the same time, which is not the case, as we can see with TV show from abroad and phone call to people living across the ocean.

3

Flat Earth Theory / Re: Sunrise and Sunset II

« on: August 10, 2017, 11:42:58 AM »

Hello,

I am very interested by the sunset on flat earth, because I don't understand how perspective works with the sun.

From this page : http://wiki.tfes.org/Distance_to_the_Sun
The sun is 2400 miles above flat earth.
At midnight in New York, the sun is above the Philippines, approximatively 8500 miles from New York (https://www.timeanddate.com/worldclock/sunearth.html?day=10&month=8&year=2017&hour=0&min=0&sec=0&n=179&ntxt=New+York&earth=0)
If I apply the same calculation that the "Distance to the Sun" page, I get an angle from the horizon to the sun seen from New York = tan^-1(2400/8500) = tan^-1(0.282352941) = 15.76 degrees

What is wrong in this reasoning?
I don't figure out because I applied the same trigonometry that the one of the "Distance to the Sun" page.
I get the same result if I draw a scale model on a page, a right triangle with adjacent side of 8.5 inches and opposit side of 2.4 inches have an angle of approx 15 degrees, so what is different with the sun, which explained how it gets below the horizon?