Visually discerning the curvature of the Earth

WHERE IS THE CURVATURE?

How many times have you seen some Flat Earther post a picture with this caption?

I've also seen a LOT of normal, educated people who post images that show a falsely curved horizon.

There is a very good paper on this very question found in this paper:

Visually discerning the curvature of the Earth

As you see in that paper basically most of us can't and photographic evidence is always going to be problematic for "proof".  The truth is you really cannot tell much from a picture without some exacting details of when and where it was taken and under what conditions, using which lens, etc.

You can make the horizon APPEAR to curve any way you want:



Wider-angle lens give more of a barrel-effect as you can see here.  This is very clearly NOT the actual curvature of the Earth - this is a wide-angle lens effect:


Of course, you could just look at the imagery from the Himawari 8 weather satellite.

There is no justified way of dismissing that evidence... Except these goofballs scream fake and conspiracy at absolutely anything and everything that disproves them -- nothing is too baseless for them.


This guy turned a month of Himawari 8 imagery into a very nice time-lapse video.

The most critical piece of information missing from most photographs is the Field of View (FOV), if we knew that and some distances we could do some calculations to find out how much curvature we SHOULD see.

Toronto CN Tower from Olcott, NY - explained

I think that the fantastic view of Toronto and the CN Tower from Olcott, NY is one of the best examples of Earth curvature you can find. It's a beautiful view over 39.5 miles of Lake Ontario so there are ample images of the phenomena.

So it is amusing that so many Flat Earthers try to use this to 'debunk' curvature by appealing to some magical version of "perspective" - not surprisingly, one for which they cannot give any formula or explanation of how we would measure such a magical effect or how it would work.

Meanwhile Refraction and Curvature are WELL understood in the surveying and scientific worlds.

Here is what the Flat Earther expects you to accept, that ~359 meters of the CN Tower are perfectly visible and in proper proportion to the actual tower - but suddenly, below that mark, 184 remaining meters just "disappear" to nothing due to their incorrect magical "perspective".  This is simply not how perspective works - if it were going to shrink the tower it would shrink it all in the same proportion.  There is no mechanism where 184 meters of a distant object would shrink to 0 pixels while the upper 359 meters remains fully visible and in the EXACT proportion we expect.

There is simply no way to excuse this as 'Perspective'.


(image credit - click to see animation)

Unfortunately, I do not have enough details for the above image -- it appears to have been taken from more than just a few meters above the water and I don't see a dock so I cannot really calculate anything specific from this image.   But what is very clear is that we can only see the tops of the highest buildings and very significant portion of the city is missing.  The image estimates it at 184m -- which seems in the right ballpark.

Here is another view where we can clearly identify the location which gives us a better estimate for the observer elevation and distance so we can do some calculations and see if this matches what is predicted by an Oblate Spheroid model.

You can clearly see here that we see just a little bit less of the Toronto skyline under these viewing conditions.


(image credit)

First of all, you have to rid yourself of the notion that 8"/mile² has ANYTHING to do with this view.

The observer is plainly NOT at "sea-level" here, so we must account for their elevation in our equation.

To that end I wrote a post: Derivation for Height of Distant Objects Obscured by Earth Curvature

That is a very long and moderately complex post so take your time.

Once you agree that the correct equation to use to estimate how much of a distant object is expected to be obscured by the curvature of the Earth is given by:

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

Then you can continue here to evaluate this evidence for yourself.  If you cannot figure out the math here then you really cannot even understand the issues or the claims so why are you even trying?

It's OK not to understand the math -- it's NOT OK to then make false statements when you have been shown, in very straight-forward and fairly simple terms.

Solving The Math

So now all we have to do is make some measurements, plug them into our formula and see if everything makes sense here.

I used Google Earth Pro to find the fairly precise coordinates for the CN Tower and the docks at Olcott, NY, those values are:

CN Tower Base (from Google Earth Pro, 2016)

(image: Google Earth Pro)

Olcott, NY Dock (from Google Earth Pro, 2016)

(image: Google Earth Pro)

Gives us a distance of 63634 meters.

And we look at the elevation profile because we need to know the "low point" and make sure we have a clear line of sight.  Our low point here is 73 meters so we will use that elevation as the BASIS for our radius.


(image: Google Earth Pro)

Next we find the Radius of the Earth, I used 43.340372 latitude and using 73 m ground-level since that is our lowest point in between: https://rechneronline.de/earth-radius/

Gives radius at our ground level (actually water in this case) of 6368.183 km

Giving us ~3.5 m for our observation elevation ([75m-73m] = 2m ground + 1.5m observation height = ~3.5m elevation) assuming they are standing on the OTHER dock.

We will first calculate the height obscured ASSUMING 0% refraction as this gives us our worst case (the greatest value that would be hidden behind the curvature) we find:

√[(d - (√h √[h + 2 R]))² + R²] - R, d=63634, h=3.5, R=6368183 = ~254.7 m obscured

And if we assume just 7% refraction we get (this means that the refraction is 7% of Earth radius):

√[(d - (√h √[h + 2 R]))² + R²] - R, d=63634, h=3.5, R=6368183*1.07 = ~236 m obscured

So somewhere between about 236 m and 255 m of Toronto should be obscured given the Oblate Spheroid model, assuming actual refraction is somewhere between 0 and 7% (NOTES: VLBL surveyors measure it using two frequencies of laser because frequency affects refraction and by measuring the differential they can get a very good estimate of the actual refraction along that sightline - EVERY sightline has a slightly different refraction, under normal conditions it will be more severe over the water/land and decrease as you angle up).

We also see that the top of the ICE East building is MISSING from the image, at 234m tall -- so a little bit greater refraction or observer elevation from here and we would expect to start seeing the top of that building -- and that's exactly what we see in the first image where the observer is either a little bit higher or perhaps refraction was a little greater that day.

EITHER WAY we are in VERY CLOSE agreement to the Oblate Spheroid model here (we are down to measurement errors) and NOWHERE EVEN CLOSE to the Flat Earth model.  On the Flat Earth model we should be seeing much, much more of Toronto below that horizon.

The distant view of Toronto only works if that water is curved, no appeal to perspective is going to excuse it, and it MATCHES our expectations when we use the correct mathematics.

(image credit)

About CN Tower: http://toronto.about.com/od/eventsattractions/f/how-tall-is-the-cn-tower.htm

Conclusion

This soundly refutes and falsifies the Flat Earth model just on the face of it and it supports the Oblate Spheroid model to within our measurement error (get a professional surveyor to get you more precise data and we can run the math and get a more precise predicted value).

Derivation for Height of Distant Objects Obscured by Earth Curvature

Correctly calculating what they expect to see if the Earth is curved is the most common error made by the Flat Earth community that I have seen. In fact, I would say they get it wrong Every Single Time (or else they wouldn't still be Flat Earthers).

The most common error made is to try to calculate the Drop Height from a tangent off the surface - this is shown below as height (D) and then use this as the value that would be "hidden" at that distance. This is just clearly wrong I find it astonishing that this is still going around.

Every time you see someone using "8 inches per mile squared" you know that they don't know what they are doing. This IS the correct formula for Drop Height (well, it's obviously not exactly 8 inches but it is surprising close, more like 8.002...) - but the problem is that it is a completely and utterly meaningless and useless calculation!

Unless the center of the lens of your eye OR camera is exactly half-way under water or buried in the dirt you are NOT viewing from height of 0, exactly at the surface. Not even if you sit your camera down on the ground. It literally has to be half-way buried/submerged.

Hopefully you can read this and understand the issue and at least stop using the wrong math.

What we want to calculate is B - and then we also want to understand the impact of atmospheric refraction, which is a vastly complex subject but we will look at the simplest case considering just one sightline and just using a conservative curvature of refraction.


(image from David Ridlen box.com - note that D should actually point toward Earth center)

Unless you have convinced yourself that measuring (D) is the wrong measurement there is really no point in proceeding - although I will show at the end that my formula and 8"/mi² give essentially the same answer (mine is more accurate) at a height of exactly 0. But if you try to use height of 0 you are almost certainly trying to use the math to lie for you.

Pythagorean theorem

Before you proceed I recommend you (re)familiarize yourself with the Pythagorean theorem which says that, given a right triangle the area of side c is the same as the sum of the area for side a + side b:


(image credit wikimedia)

Mathematically that is: a² + b² = c² where c is called the Hypotenuse (the line segment across from the right angle corner, will always the longest line segment).

We are going to use a slightly more complex version in the form of:

a² + b² = (c+δ)²

and

a² + (b+δ)² = c²

where δ is our "delta" - some unknown measure added to one of the sides that we need to calculate.

We will be using the following diagram to guide our calculations. To keep this derivation as SIMPLE as possible I'm going to solve it in two parts for you, each one using the equations above -- and then we can just substitute the first one into the second equation and get a formula we can use.

I'm also going to give you links to Wolfram|Alpha so you don't even have to understand the math very much.

Here we go...

Geometry for Height of Distant Objects Obscured by Earth Curvature

(original image)

First of all our observer is at h₀ and what we really want to know is h₁ -- this is the amount that will be obscured beyond the horizon at our total distance d₀

Think about what values on this diagram we usually KNOW.

We can see that we are working with two right triangles, and we have labeled the sides in a hopefully obvious way.

We can figure out the expected Radius (R) of the Earth using a formula given the latitude of the observation or we can use an average value if we aren't sure.

We often know about high up the observer is (h₀) - or we can make some educated guesses and then see if the calculation fits what we see. If we aren't pretty sure then we have to calculate the highest elevation we think the observer might be and the lowest -- that gives us a range of answers.

And we know the distance to the object (d₀), or we can usually get "pretty close" with tools like Google Earth. However, please note that Google Earth is giving you what is called the Great Circle Distance which is the distance along the Oblate Spheroid - straight-line distance would be less. There isn't much difference for distances like 100 miles but there is some. Just know that if you want more exact answers you have to use more exact inputs and use the straight-line distance to your object. YOU are often the source of most of the error -- do not trust your inputs to be super accurate which means any calculated values are going to have a significant error bar. Again, you need to do the calculations with a RANGE of values in mind and check to see just how much error that introduces.

That gives us...

Inputs

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 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 equation (2) h₁ = √[d₂² + R²] - R

We don't know d₂ 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 formula!

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

We arrived at this equation using nothing but simple geometry and the Pythagorean theorem so it's very easy to verify this math.

Showing match to 8"/mi² at height of exactly ZERO

Just to show that this equation simplifies to the 8.002"/mi² equation AT HEIGHT OF ZERO (0) let's test it for 6 miles at a point where the Earth radius is 6371000 meters (this is the average value). I use 8.002 because it's more accurate than the 8" approximation.

8.002 * 6² = 288.72 inches [Wolfram|Alpha] or ~24.06 feet.

√[(d - (√h √[h + 2 R]))² + R²] - R, d=9656.06, h=0, R=6371000 [Wolfram|Alpha]

Which gives us 7.31749 meters or ~24.007 feet. We have a good match here. Which gives us some additional confidence that our math is correct.

Working with the equation, examples

Ok -- but let's say we have our camera at just 1 meter above the ground?

We check our horizon distance: √h √[h + 2 R], h=1, R=6371000 = 3569.59 and that's less than d₀

√[(d - (√h √[h + 2 R]))² + R²] - R, d=9656.06, h=1, R=6371000 [Wolfram|Alpha]

That shows that only 2.9 meters will be obscured - considerably less than the wrong equation.

And at 10 meters what happens.

We find that √h √[h + 2 R], h=10, R=6371000 = 11288.1 [Wolfram|Alpha]

From this, we  know that at 10 meters we can see the distant point all the way to the base and past it, because 11288.1 is > 9656.06, we get -0.21 meters from our equation (remember that when you take a square root you don't inherently know if it is negative or positive so you have to figure that out as I have described).

√[(d - (√h √[h + 2 R]))² + R²] - R, d=9656.06, h=10, R=6371000 [Wolfram|Alpha]

We haven't even added in refraction yet. And come on, you KNOW refraction matters - you see mirages, you see light bending in water, it is clearly messing with the path of the light.

Refraction

Calculating the EXACT Refraction is complex in practice because every TINY millimeter of air can be a different temperature, density, and humidity which all affect the actual amount of refraction. And Every Single sightline from you to another distant point can have a different amount of total refraction. However, the total path integral sums up to just one number and we can express that value in terms of % of Earths radius. 0% is no refraction and we just use the equations above exactly as they are. But to account for refraction we can also add a value to the Earth's radius -- that amount can be positive or negative.

We are ONLY going to consider fairly nominal conditions where the refraction is usually between about 7% and 14% - it can go outside those values under many conditions but these are conservative boundaries.

Let's add just 7% for refraction to our last equation, this 7% is simply added to the Earth Radius to account for the refraction:

Our distance to horizon point is now √h √[h + 2 R], h=10, R=6371000*1.07 = 11676.5 m [Wolfram|Alpha]

Conclusion

Therefore, at 6 miles, from 10 meters up under nominal atmospheric conditions we expect to see roughly 11677 meters distant. Refraction here added 388.4 meters to how far we can see.

We have derived an equation that we can feel confident about using to estimate the height of a distant object that would be obscured by Earths Curvature and we can see how it varies when the various factors are changed -- and we can also see that it is very sensitive to certain values, those first few meters of elevation really make a big difference, refraction matters more over longer distances.

And we've shown that 8"/mi² is absolutely not the correct equation to be using when you want to know how much of some distant object should be obscured (given the oblate spheroid model of Earth).

In other posts I will use these equations to show that the observational evidence is actually in very good agreement with the estimates we get from this formula.

Addendum

I've noted elsewhere that many people incorrectly use the distance as given by tools such as Google Earth - But these tools all give you the distance as the length of the arc from location to location. This is NOT the correct distance to use for the above calculation.  If the distance is under 20 miles then the error is small enough that it usually doesn't matter more than a fraction of a percent.

However, for completeness, if you have the distance of the curve or arc, then the correct formula to use is the one Mick West of metabunk fame gives in his Metabunk curve calculator, which is:

formula: r/cos[d/r - asin(sqrt[h×(2×r+h)] / (r+h))] - r

Maybe I'll post a full derivation for this formula at some point.

Tools

Here is my 'GeoGebra' calculator, based on Mick West's model but I tweaked it a bit.  It reads out both the straight-line distance and curve distance so you can use either one to position the Target.

Finally I note that to calculate a better estimate for Earth radius at some latitude you can use the following calculator or formula:
Radius at given Latitude on Oblate Spheroid


There is a VERY BASIC 'calculator' for height obscured BUT, it doesn't allow you to enter the actual Earth radius at your latitude, it cannot take into account refraction, and it ONLY works at EXACTLY sea-level (cannot take into account your elevation bias) - so it gives less accurate answers.  But you can use it to get a quick estimate:

Estimate Height Obscured at h=3.5m, d=63.634km = ~254.6m

But, for example, at latitude 43.340372 where ground elevation is 73m MSL, the effective radius is ~6368.183 km and with 7% refraction, we actually should get closer to 236 meters:

√[(d - (√h √[h + 2 R]))² + R²] - R, d=63634, h=3.5, R=6368183*1.07 = 236 meters [Wolfram|Alpha]

So these things do matter somewhat, use ANY estimator with appropriate caution and awareness of your limitations to accurately measure things.  It is FAR more difficult to take accurate measurements than most people realize.  Even a few thousands of a degree in an angle, when compounded over many miles, adds up to a big error.

Flat Earth Claim: Flat Earth BUSTS Google Earth! (Prove It Yourself)

Failure Mode:  Spherical Geometry, Geometry

Here is the YouTube video in question:
Flat Earth BUSTS Google Earth! (Prove It Yourself)

One thing I want to do is remind myself and readers that perhaps this guy is really trying his best to understand things - to that end I hope my comments serve as constructive criticism.

But so many things go wrong in this short video it is difficult to know where to start.

First of all, why pick 8.5" for the base of his triangle?  That just seems arbitrary -- which is fine -- any length will work but it sure makes everything else quite confusing and difficult to work with because the sides are similar in length (but not exactly) and the angles can easily be made to look confusing.  Which I hope is what throws him off.  But if you want to make all the angles 60° you would need to make the base very short indeed, as we find out next...

Secondly, "the top angle is much larger than 60°" -- that is absolutely correct and that is because it's on a sphere where the angles simply must add up to greater than 180°.  Indeed, it is mathematically impossible for those three angles to be 60°.  The closer you make all three angles to 60° the smaller the area would be -- at exactly 60° the area would equal 0 - ZERO - Zilch - Zip - Nada - which would then make it a point and not a triangle. ONLY in ideal Euclidean (flat) space do the three angles of a triangle equal exactly 180 degrees.

Here is what it actually looks like when you draw his big triangle inside Google Earth (my end point are only approximate but in roughly the same area).  Each line is the shortest distance on the surface from point-to-point.  They "bow" out here because of perspective - which again, happens because it is a sphere.



To prove unequivocally that Google Earth is using Spherical Coordinates here let's draw a bigger triangle that is easier to work with.  We're going to go from the North Pole to 0° N/0° E to 0° N/90° E and back to the North Pole.  That gives us nice round numbers to work with and we can clearly show Google Earth is using an Oblate Spheroid model.  You can see it still looks basically the same, just a little bigger.



Google Earth gives us two key values here:

Perimeter: 30,023 km
Area: 63,758,207 km²


Now if you think about that figure for a minute you'll realize that we have two arcs from the North Pole to the Equator, each of those is therefore 1/4 of the Polar circumference of the Earth - and one Arc from 0° to 90° along the Equator which is 1/4 of the Equatorial circumference of the Earth.  This obviously makes it so much easier to do the math.

We can even feed that equation into Wolfram|Alpha and let it do the calculation (I've provided several handy direct links so you don't have to trust my math).

[(earth polar circumference in km)/2] + [(earth equatorial circumference in km)/4] = 29,989.081 km

We also find the length of each arc is:

earth equatorial circumference in km/4 = 10018.754 km

earth polar circumference in km/4 = 9985.1631855 km

That's pretty close (0.113% difference) but should it be exact?   Well, no, actually!

Google Earth uses much more exact method of calculating the distances using the WGS 84 Geodetic model (which is the reference grid for GPS as well) which gives results accurate to about half a millimeter.  But our quick swag wasn't too bad and it's certainly close enough to confirm these are not merely straight-line distances.  We amateur cartographers need to understand that when we work with values like radius and circumference these are averaged over the whole Earth.  If you added in all the corrections from WGS 84 that Google Earth does you would get the same value.  The Google Earth values are much more accurate than I need to be.

I'm just trying to break it down to a close, but not exact, model so we can understand the math here - that way we can see which one is the better fit.

Problem for the Reader: What is the straight-line distance from the Pole to a point on the Equator?
Hint: It makes a right angle with the center of the Earth and you can use the Pythagorean theorem to solve this using the polar and equatorial radius of the Earth to solve for the hypotenuse.

So we can already see the claim that Google Earth is using a "flat map" is plainly false, but it gets so much worse when you try to compute the area.

Since we know the approximate lengths of each side, let's see what we get for area of a Euclidean triangle.  For an arbitrary triangle the area is given by:

√[(a+b-c)(a-b+c)(-a+b+c)(a+b+c)]/4

You can verify that with a simple case:

√[(a+b-c)(a-b+c)(-a+b+c)(a+b+c)]/4 where a=5, b=4, c=3 => 6 -- checks out

Taking these line segments as the lengths of our triangle:

earth equatorial circumference in km/4 = 10018.754 km

earth polar circumference in km/4 = 9985.1631855 km

√[(a+b-c)(a-b+c)(-a+b+c)(a+b+c)]/4 where a=9985.1631855, b=9985.1631855, c=10018.754

Gives us 43,269,400 km² NOT EVEN CLOSE.  Treating these values as a flat area simply cannot work, you aren't even in the ballpark.  You have to treat it as a spherical section.  The formula for the area of a spherical triangle is pretty simple - we just need the angles and the radius to get an estimate (again, Google uses a much more accurate model of the exact oblate spheroid - we're going to see if a simple sphere is close enough -- this works because the oblateness of the Earth is fairly slight).

For a sphere the equation is:

ε = α + β + γ - π
A = ε r²

Which gives us 63,686,900 km² -- which again, is close to Google Earth value where flat model is FAR off.

And finally, here is what the corner looks like for each vertex of our spherical triangle - this is approximately a 90° angle -- since we have 3 vertices that adds up to 270°, clearly spherical geometry here as well.



So by every angle the flat model fails - the lengths are spherical, the area is spherical, and the angles are spherical.

The smaller you make that area the closer it will appear to be flat and the harder it will be to measure and confirm what you are seeing.  But this is only an appearance and the deviation from flat exactly matches the spherical Earth model Every Single Time.

You can download Google Earth Pro from http://www.google.com/earth/download/gep/agree.html

Quick Introduction

Will be using this space to post information that counters Flat Earth claims and share resources with others.

There is so much misinformation out there that I think that even well meaning people are being led astray.

Doubt I will change any minds (if I change your mind let me know!) but this will at least make it easier for me to state my position on various claims.

And so this little adventure begins...