Angle that the sun should be at sunset
« on: April 07, 2017, 11:52:09 AM »
Howdi, folks.

Sorry to hear about Shaq's betrayal.

Anyway, I live in Nottingham, UK, and right now as I post this the sun is somewhere over Cameroon. About eight hours from now it will be over the Pacific due south of Baja California, somewhere between the tropic of cancer and the equator, roughly 6600 miles from me. I will be well into civil twilight, going on nautical twilight. I will have watched the sun sink below the horizon, yet if it maintains a height of roughly 3000 miles above the surface of our flat earth, I should still, with my telescope, expect to find it around 27° above my eyeline.

How are we reconciling this?

Resources are https://www.timeanddate.com/worldclock/sunearth.html?iso=20170407T1932 to find out where the sun is, https://www.daftlogic.com/projects-google-maps-distance-calculator.htm to approximate distances around the world, and http://www.mathsisfun.com/algebra/trig-finding-angle-right-triangle.html because I haven't used trigonometry in nearly two decades.
« Last Edit: April 07, 2017, 02:51:02 PM by Don Lengthy »

Offline Flatout

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Re: Angle that the sun should be at sunset
« Reply #1 on: April 07, 2017, 05:44:59 PM »
You need to apply Tom Bishop's bendy light theory.

Re: Angle that the sun should be at sunset
« Reply #2 on: April 07, 2017, 08:23:45 PM »
Sweet, thanks for replying!

Unfortunately it doesn't seem like the Electromagnetic Accelerator equation is ready to be applied yet, as per your suggestion. Bishop's constant hasn't been defined and the theory isn't complete yet, let alone experimentally verified (reportedly the theory was close to complete three years ago, so hopefully the first experiments will be just around the corner!)

In any case it's very mathsy. I had to look up GCSE trigonometry to pose my question properly in the first place, so does anybody out there have a layman's answer for me?
« Last Edit: April 07, 2017, 08:25:43 PM by Don Lengthy »

Offline Novarus

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Re: Angle that the sun should be at sunset
« Reply #3 on: April 07, 2017, 09:25:48 PM »
Sweet, thanks for replying!

Unfortunately it doesn't seem like the Electromagnetic Accelerator equation is ready to be applied yet, as per your suggestion. Bishop's constant hasn't been defined and the theory isn't complete yet, let alone experimentally verified (reportedly the theory was close to complete three years ago, so hopefully the first experiments will be just around the corner!)

In any case it's very mathsy. I had to look up GCSE trigonometry to pose my question properly in the first place, so does anybody out there have a layman's answer for me?

Unfortunately that's the consequence of most flat theories - it either takes a huge amount of arithmetical acrobatics or a set of axioms that boils down to "it just is."
Sure, all systems have axioms but if the precept of the Flat Earth theory is Occam's Razor, then these mathematical twists and turns seem disingenuous to say the least.

The truth is that? Just like the positions and nature of the moon, the position of the sun is irreconcilable with the theories of the Flat Earth.
Incidentally, the bendy light theory directly contradicts another assertion by Flat Earth Theorists that noting can bend light, often used as a way to disprove gravity.
There is no one theory because no one theory can unite all the disparate facets of the various Flat Earth diaspora.
The spherical Earth model can explain it - you can't see it because it's not where the Flat model says it should be - it's behind the limb of the horizon, simple as that.