Yeah. 1,000 miles on the SURFACE of the earth should result in 123 miles of curvature on that surface.
Exactly. For example, one observer is at Palomar Observatory in California and the other is at the Leander McCormick Observatory in Virginia. The distance on the surface of the Earth is 2,160.24 miles, but a straight line through the Earth between those two points is 2,133.54 miles. Observers at those two observatories observe a difference of 0.512 degrees between the apparent position of the Moon as measured from the line running between them through the Earth.
d = b / (2 * tan (a / 2))
d = 2,133.54 / (2 * tan (0.512 / 2))
d = 2,133.54 / (2 * tan (0.256)
d = 2,133.54 / (2 * 0.0044681)
d = 2,133.54 / (2 * 0.0044681)
d = 2,133.54 / 0.0089362
d = 238,752
There's your distance to the Moon.
Observers at those two observatories observe a difference of 0.512 degrees
Let me know when you make any observations like this.
You haven't even told use which part you're disputing. Are you disputing the trigonometry, are you disputing the distance between observers, or are you disputing the angular differences in the apparent position of the Moon? Those are the three elements needed to calculate the distance to the Moon. In order for that distance to be incorrect, one or more of those things must be in dispute. It is impossible to address your dispute if you do not specify what you are disputing.
No dispute with plane trig.
The math checks out. I’m just saying, you will never measure this for yourself.
Other methods of measuring the distance to moon include laser ranging, which was done before the first (fake) Apollo missions. But can I verify this? No, because just like you can’t measure your observers angle to any accuracy, I can’t obtain a powerful laser like this. It is still an appeal to authority.
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