Offline 3DGeek

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  • Path of photon from sun location to eye at sunset?
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    • What path do the photons take from the physical location of the sun to my eye at sunset
Re: Debunking "Altered perspective"
« Reply #100 on: October 08, 2017, 02:46:51 AM »
Right.  The reason I asked is because if the sun is 3,000 miles high, it would need to be over 240,000 miles away to appear a distance above the horizon that even comes close to being equal to the size of the moon (since you provided the specs on the moon, I'll just use that). 

We're left with the sun and moon physically getting lower, or perspective causing the light to curve.
precisely. I realized later that that 3000 figure wasn't random ;D
Anyway that's the point. If you accept that light travels in straight lines, then it's basic geometry. I tried using an online trig calculator, because i'm lazy, and to have 3000 miles within a 0,02° angle the result was above 8 million miles. I won't bet my life on the figure being correct, but it looks reasonable to me.
Of course it's out of wack with reality, but that doesn't seem to have ever given pause to the good folks here ;D

I believe that the 3,000 mile height for the sun comes from the Eratosthenes experiment.  If you interpret his measurements in a flat earth frame of mind - then instead of measuring the size of the Earth, we would have been measuring the distance to the sun.

The horizontal position of the sun at sunset is easily known because the sun has to be vertically overhead some place on the ground where it's Noon.    Admittedly, the FE'ers don't have a map - so we can't measure that distance...but the fact we can fly there in an airplane sets the upper limit for that.

I went with the RE distance for that (6,000 miles) - but any other distance will do fine for the purposes of arguing this point.

Hey Tom:  What path do the photons take from the physical location of the sun to my eye at sunset?