NOT NEEDED. 8 MILES


https://dizzib.github.io/earth/curve-calc/?d0=60&h0=6&unit=imperial

This like public farting is only fun for the one doing the farting.



I just get to smell farts and wonder what's so funny.

 

Last edited: Dec 22, 2017

So when you type in 2.5ft which is 30 inches, the height of the camera, at 7.53 miles away ......both light source and camera on the frozen lake........ into the calculator you posted you get farts?
I get 20.8667 feet of missing curvature.

Fact Farts?

 

I just realized how much time I spent viewing this thread for shits n giggles. I really need to go get myself a life. But I find this thread more entertaining than I should.

 

Last edited by a moderator: Dec 22, 2017

Here's the official ISS travel pattern. It apparently doesn't just orbit the earth.

guess what happens when the crazy ISS travel pattern is laid out on a flat map like the United Nations map. I'll give you one guess...

 

the distance when measured using the map actually takes curvature into consideration so we have to use calculus. So this experiment is
flawed

 

OR U SHOULD RESEARCH

I thought ISS was fake i did not even understand why you brought this up
http://iss.astroviewer.net/

 

Last edited: Dec 22, 2017

okay lets assume the distance is 5 miles then
I still get 6.2598 ft of missing curvature.
at less than half the distance 3.5 miles I get 1.6308 ft of missing curvature

Does water have curvature but ice somehow flattens out? I want the math for that.

 

SAY THAT AGAIN but SLOWLY using CALCULUS


HERE U GOO

Well, you might start by checking your math. You are calculating the drop as 8 inches per mile squared, which is wrong. It’s wrong for a lot of reasons, but the biggest is that the 59 miles is measured along the curved surface of the Earth, not along the straight base of a Pythagorean triangle set on a plane, which only gives you a very approximate number. And it’s the wrong number anyway.

What you really need to know is how high the curvature causes water to protrude upwards between you and the far shore when you look across the lake. The answer depends very much on your viewing height, which your method also has not accounted for.


You would also need to account for parallax, and the fact the objects on the far shore appear smaller than the water at the half way point, but that’s small enough to ignore here.

When you do the math correctly, depending on the exact distance of your line of sight across the lake, it works out to a bulge of 530–540 feet, so you would expect to see only those parts of the distant skyline that protrude above that height (or a little more if we aren’t looking out right at water level):


Of course, what we actually see is…oh yeah, exactly what the math predicts.






I believe that General Relativity has nothing to do with that (quantities are to close to the Newton limit to be explained by General Relativity). We can first do a little of Mathematics to understant what should be seen 59 miles away from Chicago. Let d (9.495104m" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">9.495104m9.495104m) the distance on the surface of the Earth between you and Chicago, R (6.371106m" role="presentation" style="display: inline; line-height: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">6.371106m6.371106m) the radius of the Earth, H your altitude, and h the lowest point of chicago that can be seen. This image helps to see the relationship between all these numbers.




For example, to be able to see the buildings with an height higher than 200 m, you should be at an altitude of about 155 m. To be able to see the highest building of Chicago (442 m from Wikipedia), you should be at an altitude of about 31 m.

Now, this does not take account of any deflexion of light, for example due to the temperature gradient (see Mirage, on Wikipedia). Probably, with the lake around, in some hot days, the gradient is enough to make a deflexion so that highest buildings of Chicago can be seen, but this has to be checked by an expert.



Well… Let’s see… We KNOW the rule is 8″ x miles squared, Right?

8*59²/12 = 2320.7 ft or 707.3 m

So this MUST be correct, right? It’s math, can’t be wrong!

Well…

IF you put your eye/camera lens at sea-level, so it is half-way under the dirt/water, exactly at 0 elevation, and you ignored refraction then yes — that is what you would expect.

Now ask yourself, are those EVER the ACTUAL viewing conditions?

No - no they are not.

Since none of those assumptions are correct let’s try that again using the correct mathematics for the real-life question.

In short, you are trying to measure (D) in the image below when it is clearly (B) that you should be asking about. But for some reason, Flat Earth people can never seem to grasp this difference. (B) applies whenever the observer has ANY height AT ALL — even 1 cm. And unless you’ve stuck your camera half-way under the water you aren’t at 0 elevation (you have to measure to center of the lens).


Tom Dreyfus gave an excellent answer on the math as well, I’m just expanding on it a bit using slightly different geometry here so we can avoid doing any trig, which I think is easier for a layperson to understand, and then I’ll do the actual math for Chicago - which puts this whole nonsense to rest for the rational and sane.

Here is the correct geometry to use for an observer who hasn’t (metaphorically or literally) buried their head in the sand…


We have two right triangles to solve. This geometry is obviously correct - right? If you don’t understand this diagram then you might as well stop here. It’s also the same as the geometry in Tom Dreyfus' answer to Chicago is 59 miles from the opposite shore of Lake Michigan. Given the earth’s curvature, it should be 2320 feet below the horizon. How can it be seen? We are solving the EXACT same equation here - I’m just doing it with simple algebra.

Inputs (the values we know, to some approximation)

h₀ = elevation of observer
d₀ = total distance to distant object
R = Earth Radius (we will calculate this based on our latitude rather than using an average value)

Outputs
d₁ = observer eye-line distance to Horizon
h₁ = height of object obscured by horizon (positive when √h₀ √[h₀ + 2 R] > d₀)

Equations

The first equation we will need is to figure out the value for d₁ which is the distance to our horizon point. We will need this value in the second equation.

We can plug in these values into the Pythagorean theorem equations giving us:

(R+h₀)² = d₁² + R²

When we solve for d₁ we get (click on the equation to see Wolfram|Alpha solve it):

equation (1) d₁ = √h₀ √[h₀ + 2 R]

We can be very certain of this equation because I've provided the Wolfram|Alpha link which solves it for you.

And the FIRST thing we need to do is see if our observer can see GREATER THAN your total distance. If this horizon point (√h₀ √[h₀ + 2 R]) is GREATER THAN d₀ then your total calculation needs to be a NEGATIVE value, meaning. ONLY where √h₀ √[h₀ + 2 R] is LESS THAN d₀ then it will be positive h₁ value.

Next we just plug in the values for our second right triangle (on the right):

(R+h₁)² = d₂² + R²

And we solve for h₁, giving us (click on the equation to see Wolfram|Alpha solve it):

equation (2) h₁ = √[d₂² + R²] - R

We don't know d₂ directly but we do know that d₂ = (d₀ - d₁)

We can now combine these equations to give us our final formula (or you can calculate it yourself in the two separate steps since you need the first result anyway). You can simplify this equation further if you make a few assumptions but I don't want to do that.

h₁ = √[d₂² + R²] - R
h₁ = √[(d₀ - d₁)² + R²] - R
h₁ = √[(d₀ - [√h₀ √[h₀ + 2 R]])² + R²] - R

This is our complete formula!

Height of Distant Objects Obscured by Earth Curvature = h₁ = √[(d₀ - [√h₀ √[h₀ + 2 R]])² + R²] - R (when √h₀ √[h₀ + 2 R] is LESS THAN d₀)

Example for Chicago from Warren Dunes State Park

First we need to know what the Earth radius is at the point where we want to make our observation, we going to ESTIMATE it based on latitude for Chicago from Google Earth Pro:


There is a nifty calculator you can use here: Earth Radius by Latitude Calculator

Using 41.885367 and our elevation bias is 175m (the lowest elevation along our sightline which is the height of Lake Michigan, see more below) which gives us R of 6,368,824 meters.

The distance is ~53 miles or 85,350 meters (measured in Google Earth Pro from the shore of Warren Dunes State Park).

Observer height (182 - 175) = ~7m

Now we have to estimate the refraction… I’m going with 20% refraction as it was documented that this NOT the normal view and there was a thermal inversion over the lake. But even if you DO NOT take refraction into account (and we will do that also) you’ll just get a slightly larger value for the height obscured - this problem and the images of Chicago all still CLEARLY show curvature of the Earth! The thing you need to know about refraction here is that we can express in terms of Earth radius - so we just multiply R by our refraction factor (so 1.20 = +20%). But the person making the claim that “this is IMPOSSIBLE” needs to provide us with the EXACT value for refraction or else they cannot possibly know what is possible or not. So I’m left guessing some “reasonable” number — but since we doing upper and lower bounds here we can see any reasonable value gets us close enough to disprove a Flat Earth - the Spherical model is a better fit.

More details here: Derivation for Height of Distant Objects Obscured by Earth Curvature

Now we can plug these values into our equation:

√[(d - (√h √[h + 2 R]))² + R²] - R, d=85350, h=7, R=6368824*1.20

[Wolfram|Alpha] = ~368 meters

And again WITHOUT refraction we get a max value of:

√[(d - (√h √[h + 2 R]))² + R²] - R, d=85350, h=7, R=6368824

[Wolfram|Alpha] = ~452 meters

So we can ESTIMATE (Unless you know EVERY value to high precision you are ONLY making a rough estimate) that we would expect somewhere around 368 to 452 meters of Chicago [as measured from the water level of 175m] to be blocked at 53 miles away view from about 7m over the water, depending on the atmospheric conditions.

Since the Sears Tower is RIGHT at this limit, 442 meters tall (excluding the top spires), we would expect that under low refraction conditions we would JUST BARELY see the very tips of the tallest building (atmospheric conditions allowing) and during inversions we would see a significantly higher fraction - possibly 100+ meters.

Isn’t it AMAZING how, using the CORRECT mathematical equations and a deeper understanding of the phenomena actually being observed that we get results that align much more closely with our actual observations?

*NOTE: some of the images are from Grand Mere State Park which is 56 miles BUT you also have a higher observation point there of about +15 meters elevation (so you STILL get~368m obscured). But, please DO THE MATH YOURSELF for the EXACT observation you are testing but make sure your input values MATCH the exact picture you are testing.

What Flat Earth model cannot do is explain where the bottom of the city goes or why our view varies so greatly. Can’t be refraction because that would allow us to see MORE of the city. Can’t be perspective because that makes a distant object appear proportionally smaller in angular size, it doesn’t hide one thing behind another - if the bottom of the tower were too small to see due to perspective then the top would be too small also.

Here is the awesome photo showing a very distant Chicago skyline with the Sears Tower (aka Willis Tower) (credit to Joshua Nowicki I believe) circled in red.


And we clearly see that a significant portion of the city is below the horizon line with only tall buildings jutting up. The only question here is how much curvature are we seeing.

And we can tell there is some serious refraction going on by looking at the Sears Tower inside the red outlined area. What do you suppose that building looks like? Does it have a tall thin section below the spires on top? No, it doesn’t. So we’re seeing the top of the building Stretched by differences in the refractive index along each sight-line.

This is unequivocal evidence for the refraction.


And where do you suppose the ENTIRE bottom of the city went exactly besides hiding behind the curvature of the Earth?

Here is the Google Earth Pro elevation profile between the two showing why I used 175m as my elevation bias (it’s the height of the water which is our low-point, AKA OUR horizon is biased up by 175 meters, the lake water is NOT at sea-level but ~577′ or 175m)


And here another, more typical view where the mirage effect is inferior.

Lake Michigan mirage of Chicago skyline


~ Joshua Nowicki

Now tell me that isn’t refraction.

And another with OBVIOUS mirage effect right across the center of the ‘city’ which has significantly stretched all the buildings. This guy has some awesome images.


~ Joshua Nowicki

And a guy who did some side-by-sides comparing these photos:




LIKE I HAVE SAID WHY DONT YOU GO TO a GEOLOGIST and ask him on UNIVERSITIES?
Too scared to know the truth from an expert eh??

 

We haven't get to the good part yet

The reason you can't fly is because DENSITY
And there is the INVISIBLE FIRMAMENT which surrounds the earth
Above the FIRMAMENT, space is full of water
You can also see BOTH the SUN and MOON at night when they are on the other side of the FLAT EARTH


I think @dragonhill is embarrassed to bring it up.. lets wait and see

 

Last edited: Dec 22, 2017

Tiny font is nice too...

 

You see the problem with this discussion (and yes I know it is the the humor section) is that the facts only come from one side. One side gets to say purple unicorns mating with narwhals will produce silver dragons with golden balls... then while striking the most dashing Wharolesque pose imaginable say "prove it isn't so... dare you... can you."

That's not a discussion - not by any stretch of the imagination, humorous or otherwise.



The whole 'experiment' falls apart when she says- there is ice out here - solid ice. Then they show the image of where the torch is and she lays it on sheet of ice. For some reason I thought you were a nautical man. You can see the way the light source is distorted in the image after that, she shines down into the ice then lays back down on the ice and the lower she gets the more stable the light source becomes.

You don't need diffraction in air anymore to create a mirage - now you have water + ice + air you have

TBH I don't even mind the explaining part of this futile exercise, the part that really bothers me is having to watch those god awful videos. If you could just explain the problem mathematically and we could just have a discussion on the solution it would be quite different. However, there has yet to be one time where I watched one of those videos and didn't feel like I was dumber / sadder / more disillusioned at the world and humanity in general for having seen it.

I mean if you wanted to prove a flat earth and you lived in Utah you don't even need any of that shit lake with ice, just because you've read about thermal inversion and are like a retarded kid who stuck his dick in a light socket once - you now fear light sockets instead of just not putting your dick in them .... there are salt flats in Utah. You can use a fixed wavelength laser (the kind DJs use 10w LED emitter are fine - you can rent one for less than fifty dollars) a prism and two ladders and a two smart phone. You can do this in day light and a get pair of nautical binoculars or a decent tripod for the camera to spare us all the shaking. The prism is the cheapest and most important part, because it is how you can actually prove if the earth is flat or not after the you take a measurement of the laser from a BVR siting. The prism and the fixed wavelength laser (measure at aPE and aPR) +/- you will loose power but the only way you get a Doppler shift....

I don't need to prove anything to myself. I've been in a plane on multiple occasions, high enough to actually just turn my head and see a ball under me and black above me. I've been at sea for pleasure and for work for weeks at a time and I doubt you will find too ,many navigators who believe in a flat earth. I know not everyone gets the chance to fly above 40,000 feet, or to be at sea for extended periods of time, but that's not my fault, so w/e.

Alright - am bored once again. Also, please you have all the other dudes to talk to, you don't need to call me out by name. When I want to chip in I will, like I said I will look in from time to time, but please do me the kindness of letting me choose how much or how little to involve myself. Also, no reply doesn't mean there is "no reply," just mean there is no reply.

Cheers man - enjoy the holidays - and ffs you live in a port city - get out on the water, deep water, enjoy life.

You can actually do even cheaper on a salt flat -

rent a decent self autocollimating theodolite and some 1.5M Markers. You don't even need to go beyond visual range you can get a measure of curvature in about 10M and you can keep on going till the cows come home if you get reflector and signal posts.

Total cost under $300 for the day as long as you know how to work the equipment.

 

Last edited: Dec 22, 2017

So the image of Chicago from 8 miles out is a mirage?

 

Here's a view from high altitude balloon at close to 10,000 feet. I'm glad you see curvature at a 40,000feet just like Mythbusters at 56,000feet.

You see the curvature on the top half? That a typical Gopro trick the Mythbusters and other use with a fisheye lens to simulate curvature. The one video below is a Gopro with a non fisheye lens.

 

I think you are assuming I am following this thread a lot closer than I actually am. I remember something about the Chicago skyline at almost 60 miles out, which was a thermal inversion and a motherfucker of a nice camera - BTW the guy that took the picture 60mi out wasn't taking pictures to prove a flat earth joshua nowicki just takes pictures for a living. That has been explained to death, if you still don't get it, don't blame the explanations. Like I said you need a foundation understanding in understanding before you can even begin to understand.

I have no idea what image of Chicago from 8 miles you are referencing at at this stage in this "discussion" I have neither the interest not the will to revisit any of it.

 

Last edited: Dec 22, 2017

Now do you see the image from 8 miles out @mozee?
Oh my god, you can't see the bottom of the buildings...........oh it's only haze.

Simple question, doesn't it look like the Nowiki photo from 60 miles out?

 

Last edited by a moderator: Dec 22, 2017

No idea what you are going about or mythbusters or w/e the fk that shit is.

Been there done that, over the ocean on clear day you can see a curved horizon at about 34k feet. Didn't need any cameras or fish eye lenses, once you get to about 40-50K you can start seeing space - the atmosphere gets thin enough that the sky is just this black empty expanse.

Like I said not going to argue what I have seen or not seen... and I don't give two shits what you saw on TV or on youtube when it comes to my own personal experiences vs people who rather not have science so they can pretend its cool to have holy wars and chain up black people cause god wills it.

 

Last edited: Dec 22, 2017

You don't need diffraction on a "frozen lake" -- I don't put words in your mouth // I will refrain from doing the same -- I am not sure if you are willfully being pendantic and acting like a shitbag or if you don't get it. However, I will ask you to be more careful with my words and context and I'll do my best to be civil. Do not fucking misquote me and take my words out of context making them your words.


Yhea bro - not interested, I don't have to debunk every single fucking moron with a camera for there to be a point to it. I think we've walked far enough for today, or maybe the rest of the week. If its not Santa Klaus, the what about the tooth fairy, well about the easter bunny, or baba yaga.... and so on and so on.....

sorry man, but if hap-hazzardly stumble into something somewhat interesting like the frozen lake (cause sinusoidal diffraction along the surface of the ice/water and ray diffraction was at least different) am going to have to just zzzzzzzzzzzz on that one.

You live in the pacific northwest - go on a boat - check out the flatness of the earth for yourself. Can't cost more than 40 or 50 bucks to go out for 1/2 a day.

 

I would like to ask anyone else besides @mozee and @Medrewb if those two images above look similar. The one image shown in the bottom right of the video 8 miles out and the same skyline 60 miles out.
Not a scientific test just a first reaction.
Thanks

@mozee and @Medrewb promised they will to keep the condescending remarks to a minimum

 

Last edited by a moderator: Dec 22, 2017

You accused me of being a 'troll' a few pages back. What is your definition of a guy that keep posting BS without following the content?

The fucker took a boat from 40 miles out and took a pic 8 miles out. Is that a fucking mirage? What @Medrewb thought was curvature is only haze.OMG. All your math mumbo jumbo and you forget about haze.
I don't have to rent a boat. I watched that video of a guy that isn't asking for money to fund his research.
I think I have been cordial thru all of your BS. How about reciprocating.

 

Sorry Flat Earthers need bigger fonts sometimes. They have bad eyesight

Like example, when its night time at South America (The moon and the sun is above my country),
Americans see both the MOON and the SUN at night as the MOON and SUN has its own light.

They just don't see it sometimes