Given Hawkings measurements, yea I would think we would see more of a mirrored ball effect on the salt flats and lake bed. But clearly we don't...we see a perfect un-distorted mirror image.

Why do you think that? Have you done the math? How do you measure curvature? How do you compare it? How would you compare it to behavior on a flat plane?

What is the curvature for the Earth? How curved is that? What effect would that have on light rays?

There are mathematics to describe and work with all of these things. I can hand feed you some numbers but unless you do the work yourself to understand it you'll continue to be at the mercy of your own credulity.

Interestingly the person on the flats would also disappear in around 3 miles due to the effect of the vanishing point and the limits of naked eye human vision.

If 3 miles is the 'vanishing point' why can I see a mountain 28 miles away? Maybe you don't understand what the 'vanishing point' is? You do understand it has

*nothing *to do with things actually disappearing right? It's the point where parallel lines would converge if you extended them to infinity.

What are the limits of human eye vision? What angular size would a 6 foot person have 1.5 miles away? Or 3 miles away? How 'big' do things appear at a distance?

I get about 0.3 arc minutes for human visual acuity (and down to 1/2 arc second under ideal conditions) -- you?

angular size(α) = 2*arctan(g/r/2) -- where r is distance to thing, and g is size of thing (in the same units) - I'll use feet here... so 2*arctan(6/15840/2) = 0.0217029465258 degrees or 1.3 arc minutes.

So that seems to be about 4.3 'human eye pixels' -- So they would be tiny but you could see them. So perhaps our intuition isn't 100% perfect and we should look more carefully eh?

Of course, don't trust just the math -- test it. You'll need a decent camera where you know the exact vertical Field Of View and you can take pictures of something of a known height at different distances and compare it with the predicted values. When you find they don't match you have to try to understand why and how to test the source of error. For example, if you test carefully you will notice that the values don't agree at different distances. That's because the FOV for your camera actually changes as it focuses closer or further away -- the values given are for focus at infinity. So you need to figure out how to adjust for that.

and then you'll find that if it's not a straight line between you you are point slightly up or down - so you need to adjust for that.

And you'll keep finding little errors like that and refining your model.

But just start with shooting pretty straight at something, keep it in the middle of the image, and see what values you get. Experiment!

I'm trying to understand what my waking senses are showing me, and I do get that the globe is so huge that curves look flat, but would you agree it seems sorta odd that when we want to show a dead certain curve, its only 3 miles out, but when we see vast mirrored perfect flatness ahead (as in the above Bolivia pic) the observable curve is very very far away.

No, I calculate that you are seeing about 1.25 miles, the rest of the land which is 'most curved' relative to you, to 3 miles out fills about 1 DEGREE of angular size because it is tilted just about completely on edge to you.

Figuring this out mathematically is pretty complex but I give you the tools here:

http://flatearthinsanity.blogspot.com/2016/08/flat-earth-follies-high-altitude.html
If the curve is barely detectable at 100,000' why do you think it should be obvious when you are at just 6 feet when you are seeing a minute fraction of the Earth?

I am actually trying to understand the distinction raised above about there being two curve experiments, I appreciate your help and I think I'm beginning to understand but I'd genuinely like to know which setup method I should use to understand the curve calculator correctly.

Ok -- there are TWO different questions:

#1 how much of an object some distance away would be hidden behind the curvature if I'm at some height above the ground. Detailed math:

http://flatearthinsanity.blogspot.com/2016/07/derivation-for-height-of-distant.html
Use calculator:

https://www.metabunk.org/curve/
This version of the question is asking about BOTH triangles shown in this diagram.

If I'm at h0 on the left there -- I can see to some point on the Horizon and then PAST that to see some distant object so long as it's above h1 (on the right).

#2 If I had a perfectly STRAIGHT board that was X miles long and I placed it absolutely level with the ground at some point, how height above the Earth would I be after X miles?

To relate this to the diagram above This question says I am at d0 in the middle and I go d2 to the right, what is h1 at that point.

This is where people use 8" times miles squared.

Why 8" * d^2 is only an estimate that works under 100 miles:

http://flatearthinsanity.blogspot.com/2016/08/flat-earth-follies-how-to-derive-8-per.html
This is the LEVEL laser experiment case. We don't have an 'observer' height because we only want to know what the DELTA is between where the laser started and the height at some distance where we measure it. We don't care how high up the laser is in this case. It could 1000 feet up in the air -- 6 miles later the answer is STILL *24 feet higher than it was*. Make sense? I hope

Both of these are based on the SAME GEOMETRY but they are asking different questions about that geometry.

So again, based on the Hawking laser experiment findings can I theoretically show a person drop 6ft in 3 miles on a salt flat?

You are using the wrong math. If you have an observer AT ALL use the curve calculator and enter the observer height - it is IMPOSSIBLE for an observer, in practice, to have Zero height -- PERIOD.

Different parts of the image have DIFFERENT observer height! For light that passes through the TOP of the lens you need to measure from ground level to the TOP of the lens.

Let's say your height is 3 feet.

Put that into the Metabunk curve calculator as 3 miles and 3 feet you get:

**Horizon distance 2.12 Miles**

**Hidden Amount = 0.52 Feet**

**'Bulge' Height = 1.5**
If that was a straight slope it would be just ~0.00026801029! This is VERY VERY FLAT. But's not straight, it is a VERY SLIGHT curve that is too slight for your eye to detect because of the distance. The variations on those salt flats are greater than that but you don't complain how they make it appear not flat -- that's because even those larger variations are too small for you to see.

AND, to make it harder to see, that curvature is curving down from your visual line of sight so it's ON EDGE. The last half of it fills only about 1 degree of your vision.

Write some small words on a piece of paper -- put the paper down flat on the table and put your eye down so you are viewing the paper at a very very steep angle... can you still make out the words?

Don't you think looking at 1 mile of land on edge that is a mile away, such that it takes up just 1 degree is going to be hard to see what is going on?

https://www.geogebra.org/m/ZKqBkC4A
You can move the camera and target and zoom in & play around. I'm working on one that help visualize the horizon better but it runs VERY VERY SLOWLY for some reason.