TRUMP IN MICHIGAN: "We still got cards to play... and we know when to play them... The Space Force will be very important..."
(media.greatawakening.win)
LET'S GOOoOo
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The math is still correct for distances up to a limit. The formula begins to lose accuracy after very long distances. For the purposes of demonstrating the curve, or for figuring out whether an object should be behind the horizon, this formula is adequate.
But you are leaving out atmospheric refraction. In the radio spectrum, this has given rise to the "4/3 Earth" model, where to take refraction into account, horizon visibility is calculated as though the Earth optical radius were a third greater than the physical radius. It also occurs in the visible spectrum. When the Sun's disk is visibly touching the horizon, it is already below the physical horizon.
Even when accounting for atmospheric refraction, are there any long distance objects visible that shouldn’t be? Can refraction be used to explain away all of these objects?
The point is that you asked: "That means there will be curvature of 8” per mile squared. I ask again, are you able to demonstrate this curvature to me without using NASA as a source?"
And the answer is obviously no - because it is a false assumption about the curvature rate.
Demonstrating the curvature of the Earth is simple in a number of ways. The fact that difference constellations are either visible or not visible from different points on the Earth show that straight up is a different direction from different points on the Earth.
so flashing lights in the sky are your physical proof of curvature of earth? thats very puzzling to me. couldn't we just measure the curve? do you understand spherical geometry? can you provide me with the formula for determining the curvature rate of a 25k mile circumference sphere?
You asked for an example of how to prove the Earth is spherical. I gave you one. There are many more. I imagine you are going to find them all puzzling though. Here are a couple to think about though.
The visible horizon is further away when we stand at a greater height. If the Earth was flat, that would not be the case. You can see a ship at sea further away if you are atop a tower. Likewise for an island or land mass out at sea. It is the reason sailing ships had crow's nests.
Different parts of the Earth are in day or night at the same point in time. How do you explain that with a flat Earth model?
Gravity pulls us down towards the surface of the Earth. If the Earth was a flat plane of finite size, then as we moved away from the center, there would be larger amount of mass in one direction than in the other. This would result in a gravitational pull towards the center of the finite plane.
And onto your other questions.
Yes, we can just measure the curve.
I do understand spherical geometry. Hence why I pointed out that the formula of 8 inches / mile squared does not model a sphere, it models a parabola.
The formula for determining the curvature rate of a 25k mile circumference sphere is:
height_drop = 3980 * (1 - cos (distance * 0.0145)) Where "height_drop" is the drop in height to the observed object, 3980 is the radius of the 25k mile sphere, "distance" is the distance to the observed object, and 0.0145 is the number of degrees per mile on the surface of a 25k mile sphere.
No buddy, I didnt ask for an example proving the earth spherical. I asked you to demonstrate curvature. I ask this because I know it cant be done and its fun watching people try to explain it. What you did was start with the assumption that it is a sphere, and then used that as a basis to explain why it is you see lights in the sky in a different manner based on your position. None of that proves curvature. You chose this "example" because you are unable to directly physically demonstrate curvature.
The horizon always rises to eye level and is always flat, no matter how high you go. How do you explain the horizon rising to eye level if you are on a ball? No matter the size, on a ball, as you ascend the horizon will drop below center. For some strange reason, that doesnt happen on earth. What youre saying about horizon visibility based on height is incorrect. You need to learn about perspective and convergence. Your sight is not infinite. Ships do not sail over the curve- check the p900 videos of out of sight objects supposedly below the horizon being brought back into full view using powerful magnification.
Night and day have nothing to do with physically proving curvature. Again, you are using lights in the sky to draw conclusions.
Using a theory of a force you cannot detect also has nothing to do with curvature. Lets focus on curvature for now. I am willing to debate these other issues with you separately.
Show me the experiment that measures the curvature lol. Show me the photo that shows curvature (without using NASA or gov space agencies of course lol).
Thats a nice formula you've got there, but youre engaging in pilpul. The 8" per mile squared formula is just fine for approximating the curvature dip on distances up to several hundred miles. Only when you get past a certain point does it produce erroneous mathematical results.