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?
C Stuart Hardwick, Scifi author and science nerd.
Updated Aug 12 2017 · Upvoted by Nick Gorbikoff, Lived in or near Chicago for 18 years and Grant Hartlage, lived in Chicago (2000-2018) · Author has 8.1k answers and 47.7m answer views
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.
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Tony Miller, Software Developer
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)
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)
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
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.
~ 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:
(credit)
Watch refraction in action, you can see the buildings rise up, grow and shrink - you cannot deny we are seeing through refraction after watching these videos:
Reference Links:
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Tom Dreyfus, CEO at Redant Labs (2016-present)
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 () the distance on the surface of the Earth between you and Chicago, R () 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.
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.
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Cesar A. K. Grossmann, studied Electrical Engineering at Federal University of Santa Maria (1993)
Edit: using a calculator to find the correct numbers and in metric units, because science is done in metric units:
Cortland Richmond, Contract EMC Engineer at Belcan Engineering (2011-present)
Emmanuel Abiodun, Inventor, Investor, Entrepreneur, Engineer
Let's assume that we stand at a point on earth with our eyes at height H, at point A. Then (figure) D is the distance to the horizon as we see it from point A and R the earth radius.
From the right-angled triangle OBA we have
If our eyes are at a height of 10m from the sea level, on shore, opposite Chicago, and since earth radius is about R = 6,371,000 m, D, the horizon, is about 11.2k m. And since the Chicago is 59miles or about 84 km, you can not see the Chicago shore.
Even, if you stand at a height of 300m (a Chicago high-rise), then the horizon is about 67km and combining the two measures, in my opinion, you could still not see any Chicago building from the shore opposite Chicago.
So what you see should be attributed to refraction because of different air temperature at different levels above the lake, when applicable.
Don Dearinger, lives in Chicago
P. Everseeker McCarthy, Retired USN, Naval Engineer(Electrical), Enlisted
Marty Sinclair, Contract helicopter pilot