Claims it's fake because "it's all moving at the same rate" (unlike the guy in the car where nearby things are moving faster).
#1 he fails to account for the relative distances.
The things on the side of the road are VERY nearby and the distant clouds are thousands of times further whereas every part of the Earth is pretty far from the ISS. But still -- the horizon IS further which is why...
#2 he's just flat out wrong.
I grabbed several frames of his 'fake' video and you can CLEARLY see the effects of perspective as features are 'moving' at different angular speeds in our field of view based on their distance. This is why they get further apart.
This is an educational film intended to teach children as young as 11 years old critical thinking skills by giving them clearly false information to evaluate.
Sorry Flat Earthers, I meant to say:
AMAZING! Dr. Leo Charles Ferrari PROVES FLAT EARTH IS NONSENSE!
Perhaps you lack the critical thinking skills expected of 11 year olds?
I tweaked it a bit and added DistanceCameraToHorizon, made the Obscured object conditional (it was showing a height obscured even when Horizon point was behind the target), and added in a factor for refraction. I would rather have the Tangent line be curved but applying the factor to Earth radius is EXACTLY the same thing (just in different units). Common refraction factors used in surveying are between 7% and 14%.
Here is a another calculator that will factor in refraction and give you the distance at which you can see an object of some height. But won't calculate the height hidden at some distance:
Turns out there is a fairly straight-forward way to prove conclusively that we do not live on a flat plane. Simply pick a line of longitude (along which everyone will experience high noon at the same time, doesn't have to be one perfect line of latitude but each person should measure at high noon on the same day) and measure the length of the sun's shadow at several points of latitude from directly under the sun to as far North as you can (must have at least one measurement that is above 75°).
Requirements For Experiment
All the sticks should, of course, be very close to exactly the same length (recommend 2 to 5 meters). I used 5 meters in my example, taller is better.
The sticks should be placed absolutely plumb (straight up and down). They cannot lean in any direction.
The terrain upon which the shadow falls must be flat and level.
Everyone should be at a similar altitude, this is not as critical as length of the stick - even a few hundred meters wouldn't change the results appreciably.
Measurement must be taken when the Sun is at the Zenith for that day. Do not eyeball it, find the exact local time using a calculator such as the USNO calculator and take several measurements to find the shortest one, which will be when the Sun is at the zenith.
Ideally everyone lies along a single line of longitude and takes the measurements at the same time which eliminates any possibility that the Sun or Earth moving is causing the observed results.
Since you cannot get any of these absolutely perfect this will introduce a small amount of measurement error, so you need to be very precise if you want to perform the experiment and be able to rely on the results. The further North you can take additional measurements the less likely you will be fooled by measurement error.
ON A SPHEROID If the Earth is a spheroid then you will find that the measured shadow lengths will follow a power curve that simply cannot be accounted for on a flat plane which will be a linear function.
For example, if we had made these observations on 22, March 2016 at 12:07 UTC along 0° longitude - and a stick length of 5m (let's work in meters) we would find the following shadow lengths for each latitude.
Here are the results:
It is trivial to see here that the shadow length is NOT growing at a linear rate, it is growing exponentially. You can try it out using the PlanetCalc Shadow Length calculator
And now all you have to do is personally verify that this calculator gives you fairly accurate results for the shadow lengths (I have) and you can be certain that the plane of the Earth simply cannot be flat.
Try this simplified calculator, since it assumes sun is exactly over 0° if it isn't the Equinox you would need to compensate for that. I'm new to Geogebra so I kept it simple for now.
ON A FLAT PLANE Now let's compare the expected values above with those on a flat plane, which turns out to be trivial mathematically - we don't even need a calculator. We just need to calculate the following:
slope = (distance to sun / distance to stick) length of shadow = (height of stick / slope)
To get this formula we just have to think about two right triangles, using the slope of the first one to figure out the length of the adjacent side of the second one (height/slope).
Our first right triangle is the one from a point directly under the sun at the top of our stick, over to our next stick, say at 10° latitude, which is 1109472m for every 10°s latitude and then up to the sun, finally back down to where we started. Let's assume the sun is ~4000 miles away or 6437371m.
Since this forms a right triangle we can use "rise over run" to get the slope of our hypotenuse = Opposite/Adjacent, for example for the first 10° we have slope = j/1109462 = ~5.8
The second right triangle is the little one formed by our stick and the shadow cast by the above slope.
And the length of our shadow can be found using just height/slope! Since our height is 5m we get: 5/(6437371/1109462) = 0.86 for our first 10° jump.
Let's fill out our table with the values from a flat plane and compare. For each 10° we just multiply our Adjacent side - so we get the formula: 5/(6437371/(1109462×n))
What is happening here is that on a flat plane each addition of 10° (or 1109472m distance) gives us a CONSTANT addition to the length of our shadow of 0.861735...
You can see this by comparing any two shadow lengths:
This linear results simply cannot be fit to the actual data you get for the shadows because the surface of the Earth is curved and indeed, this proves the curvature is positive, so it cannot be concave (or the shadows would get shorter by some power function).
Here is a similar calculator for shadow Length on a Flat Earth, you can verify the distance to the sun and the position of the last stick.
Conclusion
If the data cannot fit your model then that model is falsified - the Flat Earth model here is soundly falsified by a simple measurement that is easily verifiable, even by amateurs, with very inexpensive equipment.
If you want to take some measurements at your latitude and compare them to the PlanetCalc Shadow Length calculator results, make sure you accurately capture all the factors in your video:
Date
Exact Time of measurement
Latitude
Longitude
Length of Stick (or Rod)
Flatness of the Ground
Plumbness of the Stick at time of measurement
See also 'The Noon Day Project' where children around the world have done this and logged their data. For example, I took the Fall 2008 Data and converted Latitude values to decimal form and normalized the stick size data (almost all used 100cm) and plotted the resulting shadow lengths.
Not everyone took data on the same day which introduced some error but even in this data you can see the power curve which Excel gives as: y = 11.608e0.0491x
An even simpler method?
Apparently this wasn't simple enough for some people so let me try another way to state the exact same thing.
We know from simple geometry that if we take a right isosceles triangle with an object at point (B) at some height (h) and we find the point which is at 45° (C) then we will also be a distance (d) from the point (A) and h will be equal to d at this point. So when height = distance our angle is 45° ASSUMING STRAIGHT LINES (keep this in mind). The formula for this is height(h) = distance(d) * tan(angle) and tan(45°) just happens to equal 1.
This also means the line from (B) to (C) has a slope of h/d which is obviously 1. As shown here.
Now consider point (Z) which is at twice the distance away. The slope of this line will be 0.5 and we can see this using just trivial division.
We know that SLOPE = RISE(h) / RUN(d)
So that means what we are saying is that DOUBLING THE DISTANCE is EXACTLY the same as taking ONE-HALF THE SLOPE. Mathematically:
This is a BASIC rule of fractions. And it doesn't matter what the height is or what the distance is.
And to get an angle from slope you simply take the arctan(slope) and arctan(0.5) is 26.565° You can measure the angle if you prefer. And don't take my word for it - draw it out and measure it.
NOW...
If the Earth was flat then, with the Sun directly over the equator (90° overhead) at the equinox and at an angle of 45° (slope = 1) above the horizon at latitude 45°N (which I think we all agree is accurate), with TWICE the distance at 90°N we SHOULD find the slope being 1/2 corresponding to a Sun at an angle of about 26.565° -- but this is NOT the case at all! Instead, at 90°N latitude we find the angle to sun to be 0° - lying fully on the horizon! A HUGE discrepancy with the Flat Model.
This measurement does not and cannot agree with a Flat Earth. This happens on a sphere at 1/4 the distance around. And on a sphere, the latitude of 45°N would be TILTED 45° which means ALL of our true angles point in the SAME direction - towards a distant sun as shown in this diagram.
Now, it's not really EXACTLY 45° - that is ~0.00244° off but this is too small for us to easily measure when you are staring at a blinding sun.
And if you measure the angles at other latitudes you'll find EVERY measurement is consistent with this spherical model and completely incompatible with any Flat Earth model no matter how near or far away you place the sun.
Latitude
Observed Angle Of Sun At Equinox
Distance from Equator (miles)
Which Height Is It?
10°
80°
690.3
3915.1
20°
70.01°
1380.0
3793.6
30°
60.01°
2070.3
3587.4
40°
50.01°
2760.7
3291.2
45°
45.02°
3105.2
3107.3
50°
40.02°
3450.3
2897.2
60°
30.03°
4140.0
2393.1
70°
20.04°
4829.6
1761.7
80°
10.09°
5516.5
981.7
89°
1.36°
6119.2
145.3
So we know the Flat Earth claim is FLAT WRONG just from this TRIVIAL geometry.
You can also try to calculate the height of the sun using other distances and angle measurements and you'll find that WILL NOT AGREE, you'll get a different answer for every distance and angle because THE BASE LINE IS NOT STRAIGHT, it is curved. If the Earth was Flat then EVERY angle and distance would agree and give you the same altitude for the Sun. It doesn't.
Many people are confused by Crepuscular rays because they APPEAR to converge at a point "just behind the clouds". But really they are (nearly) parallel to each other and it's just perspective that causes this appearance (as I will show).
My video below shows crepuscular rays from a rapidly moving vehicle - if the *apparent* spray of the rays was merely caused by the sun being Near to the clouds the relative angles of the rays wouldn't change as the car moves -- they would remain spraying to the left or to the right.
However, you can clearly see the "depth" of the rays as closer rays move across the field of view more rapidly than further rays and these closer rays completely flip from left-spray to right-spray. The reason for this is that the Sun is very distant, all the rays are actually parallel and are coming in at an angle towards the viewer, and because of this angle only appear to converge due to perspective effects (exactly as train tracks appear to converge in the distance but are clearly parallel). Our eyes are not used to evaluating ephemeral rays in 3-space with such complexity - but hopefully this view helps you to see and understand what the shape of that 3 dimensional volume really is (tilted rays that only appear to converge).
Here is what is going on from above:
If we could simply trace the apparent 'sunbeams' to their source then you would have to conclude from the following image that the sun goes down below the surface of the Earth.
So this is clearly a nonsensical approach to resolving what we see.
So clearly this is an absurd methodology to use to determine what we are seeing.
Do the bottoms of these buildings converge or are they parallel?
This model is from the SketchUp warehouse and I just set up a few views to take some screenshots.
I often see the objection to a spheroid Earth with an equatorial velocity of ~1040 miles per hour (or more accurately 465.10109 m/s). But these claims seem to be based on a misplaced 'intuition' that this speed would cause everything to fly off into space due to the 'centrifugal force'. However, the claimants (so far) have been unable to do the math and lack a way to conceive of what is happening, so their childhood-playground intuition fails them.
To give you a better understanding of what is happening, first we need to understand what 'centrifugal force' is, and how it is 'generated'.
First of all, there is no such force as centrifugal force! It is an apparent force that acts outward on a body that is being accelerated, but it actually arises from the inertia of mass. It is EXACTLY the same apparent force that 'throws you forward' when you slam on the breaks in your car - but when the body in question is rotating we call it centrifugal force.
Newton's first law of motion (the Law of Inertia): An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
The real force is called centripetal force and this is the force that is causing the actual acceleration. If you put a rock on the end of a string and spin it around it is the string that is applying the centripetal force to the rock which keeps it from flying away. On the Earth is it the force of Gravity that pulls you towards the center. On Earth the acceleration due to Gravity is 9.807 m/s² - this can be observed by dropping an object from various heights and observing the different amounts of time it takes to hit the ground (pick something heavy and round to minimize the effects of air resistance).
But the inertia inherient in the mass of the rock wants the rock to keep going along the path which it is currently going. The 'tug' of the string is like the breaks of your car, it pulls it around the center of rotation but anything that might be inside the rock would be pressed to the outside of the curve by inertia.
So the centrifugal acceleration is really just inertia.
What is a Force?
Force is the mass multiplied by the acceleration - for the remainder of this discussion we will focus on the acceleration component rather than the force. If you want to convert any of our accelerations into a force just multiply it by your mass in kilograms (kg). Or to make it super easy, assume a 1 kg object - which gives you kg*m/s² which is called a Newton of force.
So an acceleration of 10 m/s² acting on a 1 kg object would just be 10 Newtons of force. And on a 100kg object it would be 1000 Newtons. Which is how much force your body experiences from gravity (well, a 100kg person would experience 980.7 Newtons because the acceleration of Gravity is 9.807 m/s²).
Ok then, What is an Acceleration?
An acceleration is simply a change in the speed (which is also called velocity) of some object - if you are in a car going 1 m/s and you aren't speeding up or slowing down then you aren't accelerating.
Gravity is a field which applies an acceleration to mass upon which it acts mutually. So my mass 'tugs' on the Earth and the Earth 'tugs' on me mutually. It is from this simple observation that all of the equations of motion were first derived, it is this action that gives the planets an elliptical orbit as well. There is a great series of lectures by Richard Feynman which explain all the observations about gravity that give us our current understanding of how it works, I encourage anyone who honestly wants to understand their world better to watch them.
So what is the effective/apparent centrifugal acceleration at Earth's equator?
THIS is the question that we need to understand so we can determine if something would fly off.
If the centrifugal acceleration is greater than the centripetal acceleration then the thing would fly off.
We know that the centripetal acceleration is simply the force of Gravity (aka 9.807 m/s²).
And it turns out we can figure out the centrifugal acceleration very easily if we just know the radius around which we are rotating and something called our angular velocity.
If we were standing on the equator then our distance from the center of Earth's rotation would be 6,378,137 meters (or 3,959 miles - NOTE: the distance to the north pole is a mere 9.1 miles difference), and we know our rotational rate is once every 24 hours.
So let's define our values:
T = 86164.09054 seconds (time, 1 sidereal day)
ω = 2Ï€/T (angular velocity - simply the angle we turn divided by the time, 2Ï€ radians = 360°)
So it turns out our centrifugal acceleration is a mere ~0.0339 m/s² which is working against the acceleration of Gravity which was 9.807 m/s²
Gravity is 289 times STRONGER than the centrifugal acceleration.
So we can safely say that you would NOT fly off the Earth because of this rate of rotation.
What Is Going On Here!
Ok -- the math is a little complex and it doesn't help us understand WHAT is going on just yet. Let's look at things from a couple of different perspectives.
We remember being on the playground merry-go-round and being thrown off that and it wasn't going anywhere NEAR 1040 miles per hour!
So why is our intuition about this so wrong? The answer again lies in understanding inertia.
Do you remember standing in the middle? It was easy to sit in the middle, you didn't get thrown off, but the closer to the edge you got the greater the force you felt.
Let's use the same math from above to work out the acceleration for our little merry-go-round where we are hanging on to the outer edge (about 5 feet or 1.524 meters from our center of rotation as that is a 10 foot merry-go-round) and assume we're a 50 kg kid.
What if we SLOWLY walked the merry-go-around taking, oh say, 42.1282 seconds to go all the way around once (just to pick a random number)?
Hey! That's the SAME acceleration as Earth spinning at 1040 miles per hour! Let's see - radius of 1.524 meters = circumference 9.576 meters around... that's only 1/2 mile per hour -- that's a SLOW walk around.
What is going on here?
Both the merry-go-round and the Earth are TRYING to give us the slip -- but the Earth has a HUGE radius so even at 1040 miles per hour FORWARD (or 0.288 miles per second) and we know from the drop rate equation (8 inches times miles squared) that the drop over 1 second is about 2 inches per second or just ~0.05 m/s! Meanwhile, Gravity is sucking you down at 9.807 m/s² and since both accelerations are constant, the centrifugal acceleration really just cancels out that TINY portion of the acceleration due to Gravity, and that is at the equator -- as you move away from the equator the centrifugal acceleration drops (advanced version: to get the correct acceleration for your latitude just multiply the radius by cos²(latitude) to get your true radius of rotation, it is 1 at the equator and 0 at the pole - make sure you use latitude in the correct units; degrees or radians for your software/calculator).
So it really has to do with how fast the 'thing' is moving 'out from under you' NOT merely how fast around you are going -- the distance from the center of rotation has as much to do with it as the raw speed.
You can also try thinking about your experience in a car. Even when traveling fairly slowly, if the car takes a sharp turn you are thrown to the side with a moderate amount of force. But in a very fast moving car, moving around a larger curve of the freeway you feel much less force than the slower car turning abruptly - despite going even 10 times faster. So the size of this curve matters a lot and the curvature of the Earth is enormous, 1000 times flatter than the curve in the road.
Spin It Faster!
What happens if we spin our kids ten times faster, or once around every 4.2128 seconds?
4.2128 = 3.39 m/s² [Wolfram|alpha] - no surprise, we just went 10 x faster and got 10 x the acceleration.
And if we convert that to the force felt by a 50 kg kid we get 169.5 Newtons or about 38 pounds of force pushing our kid sideways.
That's pretty accurate.
I SAID FASTER!
Now, what if we used a motorcycle to spin our merry-go-round at say... a mere 16.5 miles per hour... You would create the Spinning Merry-Go-Round Of Death (or at least injury - don't do that kids).
I offer Flat Earthers several straight-forward challenges:
#1 Simply produce a flat map that is both area and distance accurate. This should be very straight-forward on a flat Earth.
Initial checks should be to Measure distance from Nuuk to New York (2984.16km) to Johannesburg (12834.99km) to Prague (8556.58km) back to Nuuk (4076.04km) and area of Greenland(836300 mi²), Africa (11.67 million mi²), New Zealand(103483 mi²), & Texas (268820 mi²)
If your flat map distorts any of these more than a fraction (for allowable measurement error) then it is wrong, period.
Furthermore, you can show that the values we observer on Earth only work on a spheroid - thanks to Gaussian curvature which proves you cannot flatten a spheroid into a flat map without distortions & tearing.
Distances only work to the center point and areas are distorted everywhere but exact center.
#2 show me your equations which explain the fact that we observe two (2) celestial poles (points in the sky around which all the stars appear from Earth to rotate) and accounts for the motions of all the major bodies (sun, moon, naked-eye-visible planets, and the stars where we have measured parallax such as Barnard's star, etc).
And Please Note (as they often get this wrong): there is no such thing as an exact North or South pole star - I'm NOT talking about the stars, I'm talking about the point in the sky around which they appear to rotate. Hint: Polaris is not that point, it is just pretty close.
Bonus points, explain why that celestial point moves in agreement with the axial precession of Earth's axis - which you claim isn't spinning.
#3 set up an experiment where you simultaneously measure the angle of the suns shadow off an object 5 meters tall that is plumb from at least 5 points along the same longitude line (or within ~3 degrees). 1 must be very near the equator, 1 as far North as you can get it, the remainder in between but not too close, spread out.
Taking all the permutations of angles compute the "distance to the sun" and show that all the values agree. If they diverge then the flat earth model is disproved.
If they fit an exponential (which they do) then that matches the spherical model (in other words, the further north you go, the shadows will get longer at an exponential rate)
#4 Given the standard 'spherical Earth' values for Earth mass (5.972 × 10^24 kg), equatorial radius (3,959 mi), and rotational rate (once every 24 hours), calculate the effective equatorial centrifugal acceleration at the Equator and PROVE that this would spin us off into space; that is, overcome Gravity at MSL which is ~9.807 m/s²
Hint: you are pitting ~0.0339 m/s² against 9.807 m/s² -- GO!
The profound innumeracy of the Flat Earth movement renders this one all but moot but I like to see them try to do the math because it's SO hard (rω²).
Eratosthenes is the first known person to give a good estimate for the circumference of the Earth. There are even earlier estimates but they measured poorly and got poor results.
The city of "Aswan, Egypt" lies around 24° Latitude and, in the time of Eratosthenes, there was a well in this city (called Syene at that time) and it was known that the Noon sun on the Summer Solstice would shine directly down the well, casting no shadow.
NOTE: the exact latitude for the Tropic of Cancer (the point where the Summer Solstice sun is directly overhead) changes slightly over the years. So to repeat this measurement today you would need a slightly different location along ~23°26′13.7″N (where does Flat Earth model explain that? Spherical Earth model explains it due to precession of rotating Earth).
Eratosthenes thought to measure the angle of a shadow cast in a distant city (Alexandria, Egypt) during the Solstice - and, using basic geometry, to therefore measure the curvature of the Earth between the two locations (both are within a few degrees longitude of each other).
He found the angle cast by the sun in Alexandria was about 7°14’ from true, and he estimated the distance between Syene and Alexandria at about 5000 stadia, which meant the circumference of the Earth was about 250,000 stadia (or about 50 times the distance between the two cities). This is because 7°14’ is just about 1/50th of a full circle (360°).
That distance is approximately 522 miles which means it would be about 26,100 miles around - the polar (north-south) circumference of the Earth is actually 24,874 miles so that is pretty amazing since Eratosthenes would have had to have measured the distance of 522 miles the hard way (surveying is very subject to measurement error, even little bitty tiny errors in measuring an angle can throw off the final result by many miles).
Now, there are TWO possible ways to account for the increasing angle of the shadow...
#1 the sun is far away and the angle is (mostly) a product of a spherical Earth. #2 the sun is close and the angle is just because of the position of the sun in the sky.
However, we find that Option #2 is ruled out because the angles measured at different latitudes do not agree with the linear angle model (they get stepper exponentially with distance). This is impossible on a flat earth (see details).
What about Sun Size
Earth radius is ~3959 miles. The sun is 92.96 million miles away. That is about 23,480 Earth radii away -- there is no reasonable way I can make a picture that wide.. This great distance is why we says the Sun's rays are almost parallel. But actually the Sun is also very large so those rays are not exactly parallel and in future posts we will probably touch on this as it is why shadows are fuzzy on the edges unless the object is very close to the ground. When you measure actual shadow lengths you get the results above that match a spheroid Earth with a radius of ~3959 miles.
When we look at the sun (very carefully so we don't damage our eyes) we find that the angular size of the sun (how big it LOOKS to us) is 30 arc minutes across the sky. This is true all day, the angular size of the sun in the sky does NOT get larger and smaller throughout the day (this is another way we know the flat earth model is incorrect).
If the Sun is 92.96 million miles away and has a radius of 432287.95 miles how big should it appear to us when we view it in the sky?
Well, if we made a right triangle from our viewing point to the center of the sun (92.96 million miles) then a leg out to the edge then we could solve for that angle, knowing that the tangent of the angle Θ = sun radius / distance to sun, then we just need to double that.
tan Θ = 432287.95 / 92960000
So we can take the arctan of that value to get our angle.
arctan(432287.95 / 92960000)*2 in arcminutes = 31.97' or 0.5329° (one-half degree)
This is how you compute apparent angular size of a distant object if you know how big it is.
However, since the sun is VERY far away this doesn't matter - you can enter distance = 92960000 and Size = 864575.9 and you still get 0.53288 degrees. But if you are measuring things nearby you need to be VERY accurate in your measurements.
What other experiments can you think of to do that would prove the sun is very distant?
Using your "Flat Earth" model (however far away and large or small you imagine the sun to be) how would the angular size of your 'Sun' change through a Flat Earth day?
How can you measure the actual angular size of the sun without hurting your eyes?
Since the Earth orbits the Sun in a slight ellipse can you measure the change in angular size of the sun? Does that measurement agree with the heliocentric model? (yes, yes it does)
Just because an observation or measurement agrees with your model it doesn't PROVE your model.
Your model has to account for ALL of the observations and make predictions that can be measured.
A large body of agreement with your model supports that model but it is NEVER proven. It can ALWAYS be wrong.
But a single failure of your model unequivocally disproves that model.
The Flat Earth model cannot account for why we NEVER see the bottom portion of a city at 60+ miles even when the top half is clearly visible. Perspective does NOT squish only the bottom half of something when the top and bottom are at equal distances from our eye or camera lens.
The Spherical model APPEARS to be wrong until you account for refraction and use the proper mathematics.
But what you really need to do first is watch Joshua's other video first because it shows the dynamics of what is going on during these events:
I grabbed some of the frames (in order) and put a strip side-by-side so you can see what is going on (click to enlarge).
Asking why we see what see is a good question to ask and understand, but to reach understanding you must also study all the relevant aspects. How can you watch the Chicago skyline 'rise up' from behind the horizon, constantly shimmering and flickering through all kinds of different refraction and mirage effects and deny that you seeing the effects of atmospheric refraction? You can literally see the buildings moving up and down in some cases and, in other cases, the horizon visibly drops.
Even Rowbotham cites from Britannica on how refraction curves the light in exactly this manner, he just then proceeds to dishonestly ignore it. But this shows that the greatly varied effects of refraction were clearly known in his time and that he was very clearly aware of this.
The response, however, has been everything from claims that this is NASA tricking us to just completely ignoring the evidence and claiming 'Flat Earth PROVEN!' with this picture. How can they ignore the missing half of the Chicago skyline that remains below the horizon? Well, they aren't famous for allowing facts to get in their way :) But I digress...
And for quick reference, I remind the reader that what we want to know is how much of the distant Chicago buildings should be obscured by the Earth's curvature on a spherical model. This is measure (B) from the following image, not distance (D) since the camera here is clearly not submerged half-way under the water of Lake Michigan at "0" elevation.
Note: D should actually point towards center of Earth as shown in my post on [8"×d²]
From my page above we know that:
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)
This fairly simple geometry - I didn't even use trig functions, just basic algebra - so hopefully this is fairly accessible to readers.
Grand Mere State Park, Michigan
Joshua has many (awesome) pictures and videos of Chicago from various locations along the Michigan shoreline, Warren Dunes State Park, Grand Mere State Park, and Michigan City Pier just to name a few. He is a very good photographer and you should definitely check out some of his work (facebook) (smugmug).
The image in question here was from Grand Mere State Park (source), so we will use the details for that location. The first thing we need to do is locate Grand Mere and get the geographic details for that location. Using Google Earth we find a plausible viewing location on the Dunes overlooking Lake Michigan and we get our viewing altitude (190m) and latitude 42.003712.
(aside, I welcome corrections on anything throughout this post, especially from Joshua who might have better information than my estimate of elevation and location, however, I attempted to be fairly conservative)
From here we draw a path from this location at Grand Mere to the base of the Willis Tower (I measured all the way to the base of the actual building, not just shoreline) - the upper line here is to Grand Mere, for comparison the lower (and only slightly shorter) line is to Warren Dunes.
So we're at about 90,933 meters (~56.5 miles) distance. CAUTION: there is a danger here of thinking we are "exactly 90,933 meters away", that is not true, we don't know this value exactly because we don't know exactly where Joshua was standing, so we need to keep in mind that we could be somewhat closer or further away. But we are in the ballpark sufficiently to get a good estimate of what we would expect to see, given an oblate spheroid Earth model.
Next we need to check our elevation profile to #1 verify nothing would be blocking our view (it isn't) and #2 to find the lowest-point between us, this is our elevation bias (or 'ground level'). We can't use sea-level because we have 175 meters of water and Earth between the surface of the Lake and sea-level that we can't see through. NOTE: if you use sea-level, even with zero refraction, you find only 136m hidden so we would think we could see almost all of Chicago if you use the wrong values. You really must understand what these values mean so you can measure properly and get proper results.
So we find that Lake Michigan is at about 175 meters elevation which is our elevation bias - this is the amount we need to add to the Earth's Radius to get our 'ground level' for this observation ('ground level' meaning the lowest point we could possibly 'see over'). It helps that we are viewing over a lake that is at a fairly constant level so we don't have to worry about hills in between.
We also note that we are at about ~15.5 meters over the water, plus ~2 meters for tripod, so 17.5m overall for the elevation of our observer, over 'ground level'.
Now we come to the most difficult part, estimating the refraction. We know for a fact from the video above that there are very strong refraction effects that bring Chicago all the way from being 100% obscured to being somewhat visible over the horizon. But we don't know the exact values for the moment that first picture was taken (which was ), so what we're going to do is calculate an estimate based on 0% refraction, 10% refraction, and 20% refraction and that will give us a range of values so we can compare them against our observation.
Remember that we already know from the video, that one possible observation is that Chicago skyline is completely obscured (indeed, this is the normal observed condition). We see that very clearly in the video, and then we see it 'rise up' and fluctuate quite a bit. But we never see the bottom half of the city! So this already dis-confirms the Flat Earth model, where we should be able to see all the way to the shoreline of Chicago.
Some attempt to appeal to some mythical version of perspective to explain this observation away, but perspective changes the angular size of distant objects - it simply does not shift one object behind another.
And perspective also cannot only make the bottom half of the city microscopic while leaving the top half stretched out and distorted. If you want to make such a claim you'll need to show your math and your model of physics which accounts for it. Basically this is just a non sequitur appeal to a phenomena they do not understand. Meanwhile, the science behind refraction is well-documented (if complex).
Perspective makes things appear smaller, it doesn't hide one thing behind another.
And refraction explains why we can see a little further than we think we should. I've also seen appeals to how 'perfect' this view of the Chicago skyline but I show below that it is far from perfect, there are many severe distortions. But first, let's get our best estimate of what we might expect to see under varying atmospheric conditions...
Next we use the Earth Radius by Latitude Calculator and our Latitude (42.003712) giving us our Earth radius (close estimate) at this location for ground level of 6368.78km (6368780 m).
Latitude: 42.003712 Earth radius, R = 6368780m h₀ = elevation of observer = 17.5m d₀ = total distance to distant object = 90933m
So we have everything we need here to estimate how much of distant Chicago skyline should be hidden behind the Earth. I've provided Wolfram|Alpha links which solve the exact equation I gave above with the following values (only changing refraction %):
So this gives us some estimates, under different atmospheric conditions, that estimate how much of Chicago we might expect to see at this distance and observer elevation. It varies from 364m hidden with a pretty significant (but not maximal) value for refraction, and 453m hidden at the most extreme (no refraction). I would say this pretty much matches what we see in the video and it looks like the photograph is showing just about 20% net refraction to the top of the building, with a significant amount of distortion.
Now let's consider our subject...
Chicago Skyline
The ground floor for Willis Tower is ~6 m above the water level (at 181m elevation), the tower itself is 442 meters tall up to the spires on top, and then those extend another 85 meters up. So here is a diagram showing the heights, the image is from Google Earth because it allowed me to get a straight-on shot of the building, I show an actual photo for comparison.
And a little closer up with the spire moved side-by-side showing they are approximately the same size (81 meters for the thinner segment of the building and 85 meters for the spires):
Here is an actual photo for comparison:
In the above picture, note the proportion of the Spires to the narrower top part of the building and compare those proportions again with Willis Tower in the distant image, annotated below. It is obvious on this examination that the narrow part of the building, just below the spires before the building widens, has been stretched significantly.
It isn't really possibly to be certain here because the image acuity is insufficient to identify fine details in the middle area of the buildings but it is perfectly likely that some portion of that layer is inverted and causing some of the stretching. Especially if you look at the row of buildings to the right of Willis Tower (the top line of the buildings between the green arrows), those are NOT all at a constant elevation, so you are certainly seeing extreme distortion in that layer that appears to be an inversion (and you can see it VERY clearly here).
I would say, based on this shot, that the refractive effects we are seeing vary from Superior mirage with ducting, to looming, and towering - along our different sight-lines with a narrow superior mirage just above the horizon sight-line and distortion from looming/towering above that.
This nothing even close to a clean shot of a distant skyline.
Researching this topic I also found in the 'Chicago Skyline' thread on Flat Earth Debunked someone had done a building-by-building match-up between several of the 'distant Chicago' images and a skyline shot.
You can see here that all but the tallest buildings are often obscured by the horizon and that the amount changes between this comparison image, some of the other images above, and especially in the video - where you see it change right in front of your eyes.
In the middle image (which I believe was from 30 miles away in this case) you can see the lighted Spires clearly inverted over the top of the building. and just as clearly, we can see that most of Chicago is hidden behind the horizon (due to curvature of the Earth).
Finally, here is a video with ~360 meters of Chicago gone missing:
Another objection raised is they think the buildings should be leaning WAY BACK or something.
Ok -- BASIC MATH here. 360° in a circle, 40075017m all the way around, 90933m to our target.
How are you going to detect less than 1° lean at 90km when you can't even make out any details in the HUGE buildings and they are all blurry and wavy and distorted from refraction? You've compressed an ENTIRE CITY BLOCK of width down to a dozen pixels but think you can detect a 0.8° difference that is AWAY from you. Utterly imperceptible.
No excuse for this one, someone is grasping at straws.
Flat Earthers often make the error of trying to fit one observation at a time into some distorted model of the Earth and then starting over with the next one and making up something else to explain it. None of it fits together and they do not have any actual mathematically accurate model - AT ALL.
Here are just a few of the observational facts that must be accurately accounted for in your "theory" - which I will happily review once you get it published in a reputable, peer-reviewed journal…
The tidal bulge on the Earth is not towards the moon, and not lagging behind the moon, but is out in FRONT of the moon [the moon pulls the mass of the Ocean towards it but the Earth rotates this bulge out ahead of the orbit of the moon - this gravitational bump ALSO pulls the moon towards this greater mass, accelerating the moon ever-so-slightly and causing it to move slightly, but measurably with modern technology, further away from the Earth]. http://en.wikipedia.org/wiki/Tidal_acceleration http://astro.unl.edu/classaction/animations/lunarcycles/tidesim.html [turn on all three options]
The daily apparent motion of the moon going completely around the Earth despite it having 2,413,000,000 meters to go which takes ~28 days http://en.wikipedia.org/wiki/Orbit_of_the_Moon
All of these facts are accounted for with the modern heliocentric model (relativistic) to many decimal places of accuracy based on one simple principle of mutual gravitation of mass and do not require distant galaxies to move at trillions of times the speed of light around us or other absurdities that result from myopic Flat Earther claims.
