Jeranism came out with this abomination:

And, of course, I got double-dog-dared on Twitter to explain it (or you know, asked what I thought, same thing but for adults):

@ColdDimSum @UnCastellsMes @Rjinswand @JennUndercover @FlatSlugbrains @malo_j @linxalorum @sciencegecko @mabyeez @y0unghusk @FlatRealm @bikinatroll @Erin5549 @Skeptickler @DrMozart11 What do you say about this video?https://t.co/sVlHmcwigy

— ISHAQ (@badassack) May 11, 2017

I have to say I was SHOCKED -- SHOCKED I SAY -- to find Jeranism didn't just lie about the distance. Video says 160 miles, he uses 163, but it's maybe even 166... I dunno for sure. If I want to "100% PROVE, R.I.P. Globe" I would use a video where I could authenticate all the data but I digress. It *seems* to be about 163 miles from 43°20'11.34" N 5°29'08.53" E at about 310 meters elevation. I don't know, Jeranism used 1000' "to be generous" -- give me some data I can authenticate next time. I'll take 1000' since Jeranism used it.

Oh there we go, 6:04 Jeranism cuts to different footage from much closer - that's a little bit dishonest right after showing 163 miles. Anyway...

So we're at about 1000 feet and 163 miles away.

And if you use my FEI Horizon Calculator you'll also find the same 'obscured' of 10297 feet at the target.

Same as Jeranism got with this calculator.

SURELY I'm convinced of the Flat Earth now!? Right? Irrefutable!

Um... no.

First of all we're at 43.336° north so let's fix the Earth radius... 10303.1

Crap.

6 whole feet difference, the wrong way -- but yeah, that's how much the ellipsoid makes a difference over these distances (i.e., not much).

Surely I'm doomed now...

Wait a minute. What's this? A floating mountain?

Of course, never deterred by careful observation we find them claiming this is the

**SHORELINE**!

Oh, FFS seriously?

Where does that fall on our mountain? The shoreline doesn't curve AROUND this mountain either - as I pointed out, it's ~32 miles away from the mountain itself.

Here is Canigou:

Apparently there was a global flood at this time...

So no. This isn't any shoreline. Which I very politely pointed out, along with the above picture.

But you will NEVER guess what they replied (unless you've spoke to about two Flat Earthers):

**PERSPECTIVE!**would make the shoreline - about 1000 feet BELOW US, magically RISE UP and swallow almost the whole mountain!

Of course! Why didn't *I* think of that. I'm so dumb...

As we covered before in great detail...

### Perspective (the reality based version)

At the core of perspective is a very simple mapping of a 3D point [x, y, z] into 2D [x/z, y/z]

That's literally it. It is complicated to rotate in 3D, that's where all the trig comes in, but once you have the scene ready the above transform is all you need to do for a rectilinear transform (like your eye).

We can trivially see what happens just mapping a few points into 3D.

Let's orient our 3D space like so:

x-axis = left(-)/right(+)

y-axis = up(-)/down(+)

z-axis = behind(-)/forward(+)

viewer is [x=0, y=0, z=0]

Let's just consider 4 points along x=0 from our viewer to the peak (in other words, everything is already translated and rotated into place -- we can do this because we have the relative distances and heights already).

peak at [x=0, y=8136, z=861534] (8136 feet above us (9136' total height), 861534 ahead)

below peak at [x=0, y=7136, z=858000] (1000' below peak, a little closer to us)

1000' up from base [x=0, y=0, z=850000] (z doesn't really matter here as this is along our sight line)

shore at [x=0, y=-1000, z=690340] (1000 feet below us, 690340 ahead)

Now we can simply divide. What is the result... Note: these are also slopes = RISE / RUN

These would be mapped to points on your screen. The smaller the bounding the box/clipping the smaller the Field of View. To be able to see these tiny angles we would have to zoom way in, but numerically it doesn't matter so we'll just look at the raw numbers. x will be 0 since 0/n is 0. so negative numbers will be lower, 0,0 center screen, positive numbers just keep going higher on the screen.

So just divide every y / z and we get (in descending on our "screen" order):

peak @ 0.00944362033 = 8136/861534

below @ 0.00831701631 = 7136/858000

1000' @ 0

shore @ -0.00144856157 = -1000/690340

So the shore would be nowhere near where you say -- it would be lower than 1/9th of the way up the mountain and that very clearly is NOT what we see.

ON a GLOBE Earth the only thing that changes is that our y is actually lower with distance and gets clipped by the horizon at some point. How much lower is essentially the 8" x miles^2 formula (which is only an approximation and only good for about 100 miles distance).

And that is the ACTUAL "Law of Perspective" that Flat Earthers often cite but can never actually show how it works. That is how 3D rendering software works (again, more complex because the computer has to do the rotations, texturing, shading, etc -- but that's the essence of it)

Something CLOSER and SHORTER *can* occlude something TALLER and more DISTANT -- but something below line of sight is NEVER going to magically rise up and occlude something "level" or above line of sight. That negative y number is NEVER going to be positive just because you divided it by some other positive integer (and unless you have eyes in the back of your head you aren't going to see it behind you).

Consider a peak at 7136 that is half the distance - z=425000, that's now @ 0.01679058823

That would occlude the peak at 0.00944362033. But the ground is never going to rise up to 1000' below the peak on a flat earth. All that happens is the WHOLE thing is smaller with distance -- you can either see the base of it you cannot.

R.I.P. Flat Earth.

**It's not PERSPECTIVE - it's curvature.**

But Dark Star -- 10 THOUSAND FEET OF CURVATURE! MISSING!

### Refraction (also happens in reality, ALL THE TIME)

Standard refraction for long distances is about 15% Earths' curvature for ray bending -- which makes that be about 10' of angular displacement (the Sun at horizon tends to be ~35′).

In free convection, the (adiabatic) lapse rate is about 10.6°/km or γ = 0.0106 K/m. The numerator of the formula above becomes .034 − .0106 = .0234, so the ratiokis about 1/6.6 or 0.152. In other words, the ray curvature is about 15% that of the Earth; the radius of curvature of the ray is about 6.6 times the Earth's radius. This is close to the condition of the atmosphere near the ground in the middle of the day, when most surveying is done; the value calculated is close to the values found in practical survey work. [Calculating Ray Bending - ATY]

So, using 14% puts our view just below the peaks at 8661.2 feet. Maybe it was a little higher or lower in the evening over the water? I can't possibly know -- do YOU? Do you know the lapse rate, gas composition, and exact pressure distribution over the 163 miles?

But what I can say is that it's entirely reasonable to be able to see

**that peak**from

**that observer elevation,**given what we know about atmospheric refraction.

If you want to PROVE this is impossible then you'll need to collect a whole lot of data to establish your case.

Unless you can prove that this isn't a small and common amount of refraction, and given the refraction / mirage effects we are CLEARLY witnessing here, then you have no case.

You can whine that my 'Globe is missing 2000' of curvature' but your Flat Earth is missing 8000' of mountain.

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