What the fuck is the point of this thread?

fake picture

did you even WATCH the video before making the post?

WATCH THE VIDEO JESUS CHRIST

>why are you mean?
You should start the thread by saying:
>This is my math.
>These are my references.
>What am I doing wrong?
The answer is not necessarily simple.
You take the coordinates of the three points, the photographer, and the two platforms, and figure out what the platform should look like.
The camera is to be taken into account.
"Optical errors" and whatnot are to be taken into account.

en.wikipedia.org/wiki/Horizon#Distance_to_the_horizon
>If the ground, or water, surface is colder than the air above it, a cold, dense layer of air forms close to the surface, causing light to be refracted downward as it travels, and therefore, to some extent, to go around the curvature of the Earth.

what? there is a closeup picture of the platform, and you can see everything from the closeup that you can see miles away

no camera specs/optical errors can erase the supposed 8foot/mile loss of visibility from a 1 foot elevation

its warm sea water...

warmer than the air above it? bullshit.

and yet cooler than the air above it you fucking retard

theres videos of this shit happening above hot dry land as well

provide them, then.

First question should be, can you provide a picture of those structures look like up close when that photo was taken? Without that, I have no idea what I'm supposed to be seeing.

Images at those sorts of distance have effects to account for that are extremely nuanced.

1) Local wave height. The ocean isn't actually perfectly flat. There's of course the waves you see that are a few feet in wavelength, but that's superimposed on very long wavelength waves.

2) The earth's curvature is definitely another factor to account for when looking miles away.

3) Refraction of light along the surface, depending on weather, temperature and humidity conditions. Over that long a length it's not hard to have this become a meaningful effect. In b4 mirages can't exist. Herp derp.

So without knowing what those structures look like up close, the exact wave conditions on the ocean that day, the photographers location in relation to those objects, and the refractive atmospherics, it's hard to judge whether that's the expected image or not.

>
>i know its cringe and stupid to talk about flat earth and you all jump on the circlejerk of im so smart i wont even debate flate earth

>but if this is so stupid can i just get a simple answer?
yeah. the earth can't be round or else people on the sides and bottom would fall off. we all know gravity pulls things downward.

Really guys what the FUCK is the explanation? This guy has a point; every time I see one of these threads I can't help but furiously punch myself in the forehead.

/thread

What's funny about this video is that it's by an NWO conspiracist, trying to steer his bros away from lunacy:

youtube.com/watch?v=PLdReXQ_iX4

Notice how the Toronto buildings visible across Lake Ontario change height, depending on date and weather conditions...sometimes the lower 600 feet is hidden (as the calculator predicts), sometimes you can nearly see the base of the CN Tower.

>8 feet of visibility loss per mile

Well I just did the maths on it as well. Taking the earth as a sphere we just use Pythagoras's theorem. Let [math] D [/math] be the distance from an object, [math] R_e [/math] be the radius of the earth, and [math] d [/math] be the distance the object is below the horizon, then we get: [eqn] R_e ^2 = (R_e - d)^2 +D^2 \\ \implies \sqrt { R_e ^2 - D^2 } = R_e - d \\ \text { so } d = R_e \left ( 1 - \sqrt { 1- \frac { D^2 } { R_e ^2 } } \right ) \equiv R_e (1 - \sqrt { 1 - x } ) [/eqn]
Expanding the radical about x=0, gives us: [eqn] d \approx \frac { D^2 } {2 R_e } [/eqn]

For 1 miles distance, we take [math] R_e \approx 4000 [/math] mi, then we have [eqn] d \approx \frac { 1 } { 8000 } = 1.25 \times 10^{-4} ~ \text {mi} [/eqn] Which is about 8 inches. Where the hell did you get 8 feet per mile?

Let h be the height of your perspective above the surface then the distance to the horizon
d(h)=R*arccos(R/(R+h))
At h=1 feet you get d(h)~1,23 miles.
Now let D be the distance to the object, then D-d(h) is the distance from the horizon to the object and the angle is v=(D-d(h))/R and we have that
cos(v)=R/(R+H)
where H is the height to our line of sight. Then
H(D)=R(1/cos(v)-1)=R(1/cos({D-d(h)}/R)-1)
H(9,41 miles)~59,3 feet
H(6.2)~25,74 feet

Actually, I should have called those H(0;D) since I calculated them with h=0 feet.
For a human height of ~6feet we get
H(6;9,41)~27,45 feet
H(6;6,2)~6,82 feet
Which would be consistent with the photo in OP.

What the hell are you doing here? Define your terms better, what does:
>the height of your perspective above the surface
Mean? And what's R, I thought it to be the radius of the earth, but then that doesn't work. Also provide a sketch of the geometry of the what you're doing, because it looks contrived and wrong.

It looks like you aren't focusing on the distance the object is bellow the horizon, but instead looking at the angle subtended by a person at a certain height. This is erroneous, since we want the distance of an object below the horizon, something which you won't be able to find from what you're doing. But like I said, I'd need to see a sketch of the geometry.

...

You're going to have to provide some more detail here, just don't see where you go the first part of your equation from, it looks like you're trying to do something with the angle subtended from the core of the earth, so the angle between R and R+h, but in that case we'd end up with: [eqn] \theta = \arccos \left ( \frac { R } { R + h } \right ) [/eqn], so if we wanted the length of the opposite side then we'd need to do [eqn] d(h) = R \sin \left ( \arccos \left [ \frac { R } { R + h } \right ] \right \right ) [/eqn] But I dunno, it's 5 am here, I've not slept yet, and I've already invested more time in this topic than I really wanted.

It's not a side in a triangle, it's the distance to the horizon, which is an arc.