Thanks for encouraging me to actually do the math, BeyondET. This just gets better and better.
Ok so follow along with your calculators, folks. We'll take out all the guess work, here.
Google tells us that Earth is 24,901 miles in circumference.
Half of that gives us a hemisphere or half circumference of 12,450 miles.
Now, if we want to find out how many inches the Earth curves per mile, all we have to do is take the Radius of the Earth,
Google says it is 3,959 miles,
and one quarter of the circumference,
Google says 6,225.25 miles,
and create a simple algebraic equation for the inches of drop per mile.
For this equation, all we need is the Radius distance 'IN FEET'
and the variable (X) that we are trying to deduce.
So, Earth's radius 'IN FEET' is 20,903,520 FEET.
(X) being our variable for 'inches of drop/curve per mile'
Here's our equation:
6,225.25 miles (one quarter of Earth's circumference) multiplied by (X) should equal the same amount of FEET as Earth's Radius 20,903,520.
6225.25(X) = 20,903,520
Now, if we try 8 for X, which is what 'Science' tells us is the 'inches of drop/curve' per mile of circumference on the Earth, look what we get.
6,225.25 x 8 = 49,802 inches. (Note: 'NOT' 20,903,520 inches)
Now divide that by 12 to get the feet. 49,802/12 = 4,150.16666 feet.
If our Radius is 3,959 miles, then our feet of drop should be 20,903,520 feet.
SO, this tells us that 8 inches per mile is WAY too little for all of "Science's" stated measurements of the Earth to be correct.
In order for Science to be correct about all their stated measurements of the Earth, (X) would need to equal 3,357.8603268945 inches of drop per mile.
Try it for yourself on your very own calculator.
6,225.25(3,357.8603268945) = 20,903,519.999999
If your calculations match these results, and I assure you they do unless you've done something wrong, then that means there are supposedly 3,357.86 inches of drop per mile. That's almost 280 feet or 279.821693907875 feet.
Now, for anyone math-inclined, I welcome any correction to that.
AND, having shown all that, we can see very clearly that there is absolutely no way possible that the amount of curvature this would demand would be very visible in that pic, regardless of all the compressing factors of the image, etc.
And Mt. Rainier would be 54,565.230312035 feet below visible view. That's over TEN MILES below view.
I think the dishes are done, here.
Earth is FLAT!!!