### 200 Reasons why flat-earthers are simply wrong. (Rebuke: 200 Proofs the earth is not a spinning ball.)

## 200 Reasons why flat-earthers are simply wrong. (Rebuke: 200 Proofs the earth is not a spinning ball.)

### Introduction

Proofs? Really, that's the word they went with? Oh dear. Prepare yourself for what is to come.

I'm afraid that this will be one of the longer posts. On the other hand, it'll then be out there.

**Update:**I've decided to mention this post, because it is hilarious. Apparently, some rapper tweeted about the earth being flat, when Neil deGrasse Tyson chimed in. The rapper then made a song about it. After which Neil's Nephew, also a rapper, made a rap in response.
Let us first establish the baseline. The earth is a semi-ellipsoid - think of an egg. It's hanging about in space, rotating around the common centre of gravity of several systems. The most significant ones being the sun and moon - so we're on a sort of wobbly ellipse. Note that when I say

*sphere*in the article, I usually just use it as a shortcut for discussing some shape. It's more of an ellipsoid, and then we add in geography.
How do you know this? Well, the earliest evidence is probably a disappearing sail behind the Horizon. Did they fall off? No, because Bob came back later that day with a load of fish to boot.

Amateur photo of the earth's curvature. |

*amateur*pictures and videos, too. Consider the fish-eye effect, often utilised by flat earthers to dismiss any photographic evidence of the flat earth. However, an eye-level horizon line does not get distorted; looking at a flat earth with the horizon at eye-level does not produce a fish-eye effect [http://goo.gl/wWCaHb]. There are numerous images showing that there is a pronounced effect for an eye-level horizon, thus directly providing evidence against the flat earth.

Let's move on. From mathematics, we can use the distances between known points - say, capital cities - to show that this can't be done on a flat surface. (See also: Natural History: The origins of life, the universe and everything!).

Another good one is that the path between capital cities (or other references) on a flat map aren't lines - they are great circles. You can calculate this by looking at the minimum of an integral that defines the travelled distance. It's done in a subject called classical mechanics. That's the stuff that explains why you know the name

*Newton.*

*I feel inclined to warn you; this will be*

*a long post*. Not in the least because I will, at times, be quite thorough in explaining how things work. For instance, the article explains navigation by using a sextant and how lift is generated for an airplane. The following meme seems applicable.

A meme of Tony Stark and Dr. Bruce Banner. It's from the end of the first Avengers movie. |

### Mechanics

I was wondering what I'd share on classical mechanics here. It seems pertinent to show a movie from my teacher, who took to uploading youtube videos last year. However, which one? I've decided to link to the one on variational calculus. So what will I include? Well, gravity seems very pertinent. So let's include the movie on Kepler's Laws - which ends with a rather brilliant python simulation. Both of these movies are by Dr. J. M. Thijssen of Delft University of Technology.I do hope you find the math accessible enough. If you don't, then look at the movie's ending, which explains why it is easy to see if Kepler was wrong. Note that this Lagrangian formalism, as it is called, is really just minimising the energy (or more accurately, the action).### 1. The horizon always appears perfectly flat.

I just gave you an amateur image that shows it doesn't. In fact, when I was flying to Geneva, I did see the curvature. Here is another one.

*Update:*A forum topic discussing this had a comment that this is in direct conflict with what Neil said. Surprisingly, being at $10$ km+ in the air or higher (amateur image uses a floating object) does not qualify as

*crawling*upon the earth. As you will see in this article, when we discuss a paper from Applied Optics, 10 km is high enough to begin discerning the curvature.

### 2. The horizon always rises.

How to calculate the distance to the Horizon given your height. |

No, it doesn't. What do you think the horizon is, some mythical creature? Let's show what it is. Suppose for a moment that you're shining a laser while standing on a huge sphere. You start by pointing it upward, then move it downward towards your feet. The Horizon will be that point where the laser first hits the sphere. Yes, again, it is that simple. A slightly more complicated story allows us to calculate. I took it off Wikipedia/Horizon, and it includes the calculation.

### 3. The natural physics of water.

The appropriate physics here are just minimising the gravitational energy. If the earth is a sphere, then that just means going to the lowest point. Which means that, if we do away with geography for a bit, we're just talking about a uniform distance from the surface. In that case, level plains would exist in exactly the same way as the earth is flat - i.e. they wouldn't, but to our eyes, on the scale we evolved to see, it would apparently look flat.

### 4. Rivers run down to sea-level.

*If the earth were a ball then the mississippi would have to ascend 11 miles.*What the hell is up on a sphere? The direction perpendicular to the surface is simply

*outward*. Which means that, with regards to that line, there's no climbing. There's only a simplistic drawing going on here. It's a sphere, it has rotational symmetry. So, your 11 miles of climbing is someone else's 11 miles of gently flowing downwards.

### 5. One portion of the Nile river.

Here's more confusion about what

*up*means. Apparently, the writer thinks that if you don't glue them down, things in Australia would fall off if they aren't glued down. Because, you know, gravity is a myth or something?### 6. Basically, measurements of the curvature.

Flat earthers, for some reason, like this story about the Bedford level experiments. Too bad it is a lie.

As the surface of the water was assumed to be level, the discovery that the middle pole, when viewed carefully through a theodolite, was almost three feet (0.91 m) higher than the poles at each end was finally accepted as a new proof that the surface of the earth was indeed curved. [Wikipedia]

### 7. Surveyors, etc., are never required to factor in the curvature.

Aren't they? Are you absolutely sure? Really now? I'm guessing that by now, you realise that land surveys covering huge areas of land will use a method that take into account the curvature of the Earth, which enables the accuracy of the land survey to be even more detailed. And so on, and so forth.

### 8. Suez Canal

It's literally the instruction 'dig this deep'. Can you make an incision, 1 cm deep, in an apple?

You can? Well,

You can? Well,

*therefore that apple is flat.*Well done, that person.### 9. Some random quote.

An unknown person, claimed to be an engineer, is claimed to have made a quote. None of this with any citation or whatsoever.

So, let's go back to the surveyors (7). These guys here are the ones that tell you what the distance is. And they.. account for the curvature. Look in the second link, the table; for 60 km you're talking about a 243m correction.

Who do you think sets out the lines and similar required things for the construction workers? Surveyors, again.

Basically, what is said here in this quote is something like this.

Basically, what is said here in this quote is something like this.

*I am Mr. Winckler, and I used to design large-distance connections. I never really bothered with checking the data provided by surveyors, and never really figured out that they had already accounted for the curvature. Also, I heard this thing about squaring circles or not doing it, anyway, the point is that it sounds fancy.*### 10. Time to confuse stuff again.

I'm not even going to bother checking the numbers. It basically says, the connection from A through B to C is really long, so B will be really high up but it's not.

What is defined as height? Probably something with a reference. So, thinking in spherical coordinates (longitude, latitude and distance from centre), a height of 2 meters really means something along the lines of

What is defined as height? Probably something with a reference. So, thinking in spherical coordinates (longitude, latitude and distance from centre), a height of 2 meters really means something along the lines of

*"at longitude, latitude the distance from the centre is the average distance + 2 m".*

Apparently, this is difficult to understand? I provided you with an image, too (Click to enlarge).

### 11. Another civil engineer who is wrong.

I quote: "

*taking one station with another over England and Scotland, it may be stated that all stations are on the same relative level".*And there you have it - the picture in #10 is valid! Flat earthers apparently don't understand how height works.### 12. It is not practise in laying out public works to make allowance for the curvature of the Earth.

As shown before, in #7, surveyors do take the curvature into account. So there's that. More people that don't understand where their data is coming from. I'm afraid this sort of thing only gets worse with computers - people genuinely don't understand where things are coming from. It's like the urban myth of children that don't realise milk comes from cows.

### 13. Let's not understand what the Horizon is for somewhat longer.

I'm beginning to think this is really

*100 different ways of explaining we don't know what the Horizon is, followed by 100 different people saying they don't understand what surveyors give them.*

In the previous blog post on flat earth, I provided a pretty clear picture to explain where they went wrong. The horizon line is not the tangent line. You grab your height and look for that line which just touches the surface. The point where it touches is the horizon; things at the same "height" (i.e. radius + height from surface) will be hidden behind it.

### 14. Another way of saying they don't understand what the horizon is.

See approximately all previous points.

### 15. Airplanes are apparently not subject to gravity.

They claim that airplanes would have to

*dip their nose down*or fly off into space. Watch the movie at the start on the Kepler problem. Come on guys, it's just an orbit! It's literally something done in high school. Let me show you.
\[G \frac{m_1 m_2}{r^2} = \frac{m_1 v^2}{r} \\ G \frac{m_2}{r} = v^2 \\ r = \frac{G m_2}{v^2} \]

That was Newton's second law for the net force being simply gravity and the centrifugal force. Yes, the airplane has to dip their nose down. It does so because of gravity. And the downward acceleration is

In the earlier days of Newtonian physics, it was demonstrated that light interfered. Therefore, it was concluded that light was a wave and not a particle. Today, we know it is both.

Either way, if it is a wave - especially coupled with Maxwell's equations that showed light is an electromagnetic wave, what is it that is going "up and down". What is the medium in which the light propagated?

Well, for a long time it was thought (and taught) that there was something called the æther. Because light moves in this medium, it doesn't act the same as the other physics do when you switch between two reference frames. (Reference frames are like standing still, or sitting in a car at constant speed).

From wikipedia, we learn that one hypothesis was the following:

That's right. He did not confirm the hypothesis of Aether Drag, in particular the one of Klinkerfeus. More generally, he did not confirm any hypothesis.

Some models allowed for this - for instance, wikipedia mentions Fresnel's model. The question was finally resolved, in 1887 by the Michelson-Morley experiment. Ultimately, this led to the Theory of Special Relativity by Albert Einstein.

I think we got sidetracked. Airy's failure - a term only used by flat earthers, as far as I can find - refers to an experiment that falsified a specific model and generally any aether model that included first-order drag. There is nothing in the experiment about relative motions and so forth. So, it's a false claim (also known as a lie).

The common explanation, at the time, for his failing to measure the parallax is that these stars were

*exactly*so that you your rate of falling equals the earth's curving away from you. Honestly, this was figured out hundreds of years ago.### 16. Airy's "failure"

In science, one often not only tests the primary hypothesis but also a number of secondary hypotheses. This particular primary hypothesis was about the aether drag, due to the earth's motion. Let's put some quotes down, from the actual paper:

Professor Klinkerfues maintained that, as a necessary result of the Undulatory theory, the amount of aberration would be increased, in accordance with a formula which he has given, and he supported it by the following experiment.

A result of physical character so important, and resting on the respectable authority of Professor Klinkerfeus, merited and indeed required further examination.

Had the result of the observations been confined to the determination of an astronomical constant, or the variation of its value for different telescopes, I should not have thought it worthy of communication to the Royal Society. But it really is a result of great physical importance, not only affecting the computation of the velocity of light, but also influencing the whole treatment of the Undulatory Theory of Light.

It is perhaps not clear what is happening here. I'll attempt to explain.Remarking that the mean results for Geographical Latitude of the Instrument (determined from observations made when the Aberration of the star had respectively its largest + value and it largest - value), agree within a fraction of a second, I think that the hypothesis of Professor Klinkerfues is untenable.

In the earlier days of Newtonian physics, it was demonstrated that light interfered. Therefore, it was concluded that light was a wave and not a particle. Today, we know it is both.

Either way, if it is a wave - especially coupled with Maxwell's equations that showed light is an electromagnetic wave, what is it that is going "up and down". What is the medium in which the light propagated?

Well, for a long time it was thought (and taught) that there was something called the æther. Because light moves in this medium, it doesn't act the same as the other physics do when you switch between two reference frames. (Reference frames are like standing still, or sitting in a car at constant speed).

From wikipedia, we learn that one hypothesis was the following:

So what was the conclusion? On Airy's own wikipedia page, it is summarised concisely:Thus at any point there should be one special coordinate system, "at rest relative to the aether". Maxwell noted in the late 1870s that detecting motion relative to this aether should be easy enough—light travelling along with the motion of the Earth would have a different speed than light travelling backward, as they would both be moving against the unmoving aether. Even if the aether had an overall universal flow, changes in position during the day/night cycle, or over the span of seasons, should allow the drift to be detected.

Like in all other aether drift experiments, he obtained a negative result.

That's right. He did not confirm the hypothesis of Aether Drag, in particular the one of Klinkerfeus. More generally, he did not confirm any hypothesis.

Some models allowed for this - for instance, wikipedia mentions Fresnel's model. The question was finally resolved, in 1887 by the Michelson-Morley experiment. Ultimately, this led to the Theory of Special Relativity by Albert Einstein.

I think we got sidetracked. Airy's failure - a term only used by flat earthers, as far as I can find - refers to an experiment that falsified a specific model and generally any aether model that included first-order drag. There is nothing in the experiment about relative motions and so forth. So, it's a false claim (also known as a lie).

### 17. If you are a flat earther, lights remain at constant intensity no matter the distance.

Actually, that's it. Inverse square laws. Although we also have to think about spectra, in particular of stars with spectra outside the visible spectrum of light. Also, the cosmic red shift due to the expansion of space will put (really, really) far-away objects outside of the visible spectrum anyway.

### 18. The Michelson-Morley and Sagnac experiments attempted to measure the change in speed of light due to Earth's assumed motion through space.

Actually, Earth's motion through space was already a long-established fact. The non-flat earth can be dated back to Ancient Greece, and any sailor that did understand what a Horizon was, contrary to flat-earthers, knew it anyway.

As we've just explained, the experiments tried to confirm to aether drag - and failed.

### 19. 400 Years ago, someone hypothesised something and turned out to be wrong.

Stellar parallax is actually a thing - and it is measurable. Indeed, Wikipedia notes:

This is actually a smaller parallax than the error Brahe made himself:This parallax does exist, but is so small it was not detected until 1838, when Friedrich Bessel discovered a parallax of 0.314 arcseconds of the star 61 Cygni.

Yes, the parallax is 0.314 arcseconds versus Brahe's systematic error of 2 arcminutes.The median errors for the stellar positions in his final published catalog were about 1'.5, indicating that only half of the entries were more accurate than that, with an overall mean error in each coordinate of around 2' .

The common explanation, at the time, for his failing to measure the parallax is that these stars were

*really*far away. He then reasoned that, if this was so, that the distance to these stars would be over 600 times that of the sun to saturn. Well, the closest star is at 4.37 light years, while the distance from the sun to saturn is 0.0001515 light years. I think that we can, in fact, agree that $ \frac{4.37}{1.515\cdot10^{-4}}\approx 3 \cdot 10^4 \gg 600 $.### 20. Flat-earthers have never heard of the conservation of momentum.

And here we go again. A cannonball that is within the cannon is moving with the earth's rotation, yes? Good. So it has momentum, namely $p_0$.

Then, we fire it. This introduces some force; $p(t) = p_0 + \int_0^t\:dt\:F(t)$. So now, it doesn't fall significantly to west because it is already moving with that speed. In fact, so is the air - and most other things.

Then, we fire it. This introduces some force; $p(t) = p_0 + \int_0^t\:dt\:F(t)$. So now, it doesn't fall significantly to west because it is already moving with that speed. In fact, so is the air - and most other things.

Was this tested? Of course. You remember about the scientific method, right?

In conclusion: Yes, conservation of momentum is a thing.Hooke, following a 1679 suggestion from Newton, tried unsuccessfully to verify the predicted eastward deviation of a body dropped from a height of 8.2 meters, but definitive results were only obtained later, in the late 18th and early 19th century, by Giovanni Battista Guglielmini in Bologna, Johann Friedrich Benzenberg in Hamburg and Ferdinand Reich in Freiberg, using taller towers and carefully released weights.[n 1] A ball dropped from a height of 158.5 m (520 ft) departed by 27.4 mm (1.08 in) from the vertical compared with a calculated value of 28.1 mm (1.11 in).

### 21. If the earth was spinning, flying vehicles could hover and wait for their destination to come to them.

Actually a fairly interesting one - I've heard it before from sincere questioners. Suppose, for a moment, that the question is a fair one.

What would this mean? It would mean that the wind isn't rotating with the earth. The speed they mention is

*``over 1000 mph''*. If the air isn't rotating with the earth, then it follows that the wind speed is always*over 1000 mph*(wind is relative motion between you and the air)*.*It's not, so we deduce that the air rotates with the earth.
So, how much force would that generate on a vehicle flying? The formula for the drag force due to the air is $F_D=\frac{1}{2} \rho v^2 C_D A \approx v^2 A$. So the force is about $10^6\:\left[\frac{N}{m^2}\right]$. I think you'll agree that the $A$, the cross sectional area of a helicopter or a hot-air balloon or anything flying is over $1\:\left[m^2\right]$. So the force is over $10^6\:\left[N\right]$. A Boeing 747-200B weights about $4\cdot10^5\:[\text{kg}]$ - so it would accelerate at more than $2.5\:\left[\frac{m}{s^2}\right]$. It will take less than 8 minutes to get to the

*over 1000 mph*rotation speed.From this book, titled `Spacecraft systems engineering', I quote:

The owner of that book adds that this is a very good approximation at low altitude, such as for airplanes, but gets worse if you increase your altitude. Whether or not this is gradual or not we don't know. I could imagine that, say, entering the troposphere causes a change in the rate at which it changes.Since the Earth's atmosphere rotates approximately synchronously with the Earth, in general the drag force has a component perpendicular to the orbit plane.

Another point to make is that aircraft define their position relative to an air traffic control tower - which is rotating with the earth. So, for aircraft,

*hovering*or equivalently keeping equal distance to the control tower means rotating with the earth.### 22. Flat-earthers still don't get the point.

Felix Baumgartner jumped from a capsule that was suspended from a balloon. Not surprisingly, the balloon was already in turning with the earth, albeit not in geosynchronous orbit. The conservation of momentum strikes again.

### 23. Gravity magically and inexplicably drags the atmosphere in perfect synchronization with the earth.

It is not really magically, it's called drag (drag is sort of like `friction' but with fluids instead of solids).

Let's first address their claim that

*rain, fireworks, etc. would behave differently if both the earth and the atmosphere were spinning*. This is simply false; they wouldn't. Let's go to Wikipedia/Centrifugal Force. We find:
\[\underline{F}-m\frac{d\underline{\omega}}{dt} - 2 m \underline{\omega} \times \frac{d\underline{r}}{dt} - m\underline{\omega} \times \left(\underline{\omega}\times\underline{r}\right]=m\frac{d^2 \underline{r}}{dt^2}\]

You can see four terms there. The $\underline{F}$ on the left and the term on the right are the ones you are familiar with. Everything else is what happens due to

*spinning*of the reference frame.
The first term is the Euler force, which is about a spinning system that is spinning at a changing rate. For the earth, which is roughly spinning at a constant speed, this contribution is zero.

The second term is the Coriolis force, which depends on the rate of spinning and the velocity of the object. You probably won't be surprised to learn that the earth spins very slowly; only 360$^o$ or $2\pi$ per day or $\frac{2\pi}{86400} \approx 7.27\cdot10^{-5}\:\left[\frac{rad}{s}\right]$.

Finally, there's the centrifugal force. It's cancelled by gravity; after all, the frame is rotating because of gravity. Not exactly; in fact, gravity pulls you down which causes friction with the rotating surface which is what attaches you to the rotating frame.

That's actually a good point to go back to the magic, also known as

*friction or drag.*What's the difference between water, honey and syrup? Put a spoon and try to move it. That's viscosity; resistance to movement in a fluid. In essence, the particles moved by your spoon bounce off the other particles and back, causing a resistance. For the atmosphere, the driving force is the friction with the earth's surface. Whenever the lowest `layer' slows down, this friction will cause it to sync speeds again. Then, the next layer has friction with the lowest one; and so on, and so forth. Because the `outer layer' is exposed to space - which is, essentially, vacuum - there is no force that slows it down. The atmosphere will move at (roughly) the same speed.
Anyway, this 'layer has friction with layer' type of explanation works sort of nicely. The viscosity is a material property that we can use in a few specific models to properly calculate this sort of stuff.

Let's move on with this train of thought, and think of a large sink filled with water. Grab a tea spoon, and stir it. If you do it with your hands, you have a vortex pretty quickly. But with a teaspoon, it takes a long time - but at some point, all of the fluid will move.

The earth has been spinning for a few billion years. The entire atmosphere moves with it. There's no magic, just some physics. The sort of physics you can do with a teaspoon, a sink, a water tap and some syrup.

Let's move on with this train of thought, and think of a large sink filled with water. Grab a tea spoon, and stir it. If you do it with your hands, you have a vortex pretty quickly. But with a teaspoon, it takes a long time - but at some point, all of the fluid will move.

The earth has been spinning for a few billion years. The entire atmosphere moves with it. There's no magic, just some physics. The sort of physics you can do with a teaspoon, a sink, a water tap and some syrup.

### 24. Cannons conserve momentum, too.

The argument here is, like before, that cannonballs don't conserve momentum for unclear reasons. As far as I know, they do and have always done so.

Moving on.

### 25. Planes move relative to a control tower.

This is similar to #21. The earth and the atmosphere spin. The planes velocity is taken relative to a point on the surface; so flying $500\:\left[\frac{m}{s}\right]$ is relative to the rotating control tower.

### 26. For some reason, exactly the same as 25.

That's right; it's exactly the same misconception as #25.

### 27. A similar misconception, now with landing runways.

Here, the misconception is expressed in airplane runways.

Again, the reference point for an airplane's speed is a point that rotates with the earth. As a result, it is synchronised with the earth's rotation and therefore the runway is in the same reference frame as the airplane.

Again, the reference point for an airplane's speed is a point that rotates with the earth. As a result, it is synchronised with the earth's rotation and therefore the runway is in the same reference frame as the airplane.

### 28. Clouds are in the atmosphere, too.

Yes, clouds are in the atmosphere; they are part of it. So they move with the rotation as well.

I wonder if they're ever going to make a different point.

### 29. If the earth and atmosphere constantly spin, this would be measurable.

You can't see it; you are moving with it. You can't hear it; hearing is sound, and sound is a pressure wave in that spinning atmosphere. You can measure it [Wikipedia/Earth's Rotation/Empirical tests:

- Newton predicted that the earth's rotation would flatten the poles and bulge the equator. This was found in the 1730s.
- The differing rotation of cyclones in the Northern and Southern hemispheres are due to the coriolis force, which is due to the earth spinning.
- The eastward deviation of a body dropped from a height, suggested by Newton in 1679, was independently measured by Guglielmini, Benzenberg and Reich. For instance, a ball dropped from 158.5m departs 24.7mm from the vertical.
- The most celebrated test is the Foucault pendulum, built in 1851.

A rather nice leaflet on the Coriolis force can be found on the Grenoble Institute of Technology website.

As you can see, it

Elements of airplane performance, section 1.3 starts with deriving the coriolis and centripetal forces. Assuming some things, they calculate that the centripetal contribution to the

*is*measurable and*has been measured*. For instance, airplanes have to account for it.Elements of airplane performance, section 1.3 starts with deriving the coriolis and centripetal forces. Assuming some things, they calculate that the centripetal contribution to the

*magnitude*of the acceleration is $a_t = \omega_e^2 R_e \approx 0.034\:\left[\frac{m}{s^2}\right]$, where the most important contributions are being at the equator plane and close to the earth. As for the coriolis contribution, for a body moving with a velocity of $555\:\left[\frac{m}{s}\right]$ they calculate a contribution of $a_C = 2 \omega_e V \approx 0.081 \:\left[\frac{m}{s^2}\right]$. Keeping in mind that planes weigh thousands of kilogrammes, we calculate that the contributions to the force (in Newton) are of the order of tens of Newtons. As a result, it is largely neglectable.Avionics Navigation Systems, Chapter 2 is

*`The Navigation Equations'.*Section 2.2 is

*`Geometry of the Earth'*, and immediately starts describing the oblate spheroid or ellipsoid. It notes the

*equatorial radius*$a$ and

*polar radius*$b$, the eccentricity and the flattening. They also explain how to get the meridian and prime radii of curvature, which are important for so called dead-reckoning navigation. The latter, simply put, is "I went this far in this direction, so I should be here". Then, it lists a number of

*Coordinate Frames*in section 2.3, which I will also list (but with my explanation).

- Earth-Centred, Earth Fixed. Basically this is putting the common xyz axes in the middle of the earth, so that the x-axis goes through the Greenwich Meridian, the z axis through the North Pole and the remaining y-axis is perpendicular to these.
- Earth-Centred Inertial. Here, the centre of the coordinate system is the earth's centre of mass. However, only one axis points along a familiar direction, towards the North Pole. The other two point towards far-away stars.
- Geodetic Spherical Coordinates. Simply put, longitude, latitude and the altitude.
- Geodetic Wander Azimuth. These are local coordinates. One axis is chosen upwards, and the other two lie in the plane. I believe the common definition is to sort of make one point towards the north, but if it's not a good match, then so be it. These are a bit vague, I feel.
- Direction Cosines. Some weird definition that is relative to the first frame.

The list goes on, actually, but what I found important is to give you a strong sense of how strongly avionics has to account for the earth being spherical. If they didn't have to, they wouldn't. Navigation in particular is extremely sensitive to this, and we'll talk about it more in #34.

### 30. Wind going in different directions at different heights.

The claim that this is consistent only with a non-spinning earth is, of course, not backed up by anything.

We now known that, roughly, the entire atmosphere spins with the earth. So, if a stationary earth is compatible with this, then so is the earth in a rotating reference frame.

We now known that, roughly, the entire atmosphere spins with the earth. So, if a stationary earth is compatible with this, then so is the earth in a rotating reference frame.

It's actually taking quite a bit to find a nice source on the simple fact that the wind can have different directions at different heights, and that this is not a problem. The simplest demonstration is taking two regular fans, and putting one on the table. Make the air flow parallel, and there's no problem for making them pass on different heights.

I did finally find one! Here, you find some time lapse videos that show clearly

*that*it happens. This explains how it works. Here's another thing that explains.
The short conclusion is that volumes of air of different temperature and water vapor content (high is a cloud) can move in different directions.

It's not too fancy, and as I explained it doesn't falsify the earth's spin.

It's not too fancy, and as I explained it doesn't falsify the earth's spin.

### 31. Apparently, clouds can't exist.

A hippo crate, since hypocrite is, indeed, hard to spell. |

So, here they claim the following:

Suddenly, it's not magic anymore but necessity. There's nothing related to the picture on the right there.If the earth-globe rotates on its axis at the terrific rate of 1000 miles an hour, such an immense mass would of necessity cause a tremendous rush of wind in the space it occupied. The wind would go all one way, and anything like clouds which got `within the sphere of influence' of the rotating sphere, would have to go the same way.

Well, we've already thought about this - but we were looking at the bulk movement of the atmosphere, that it had to spin with the earth.

However, the air isn't uniform. We already considered this for clouds; there's differences in density, amount of moisture, temperature and so on. Pressure differences also happen - they are related to the local weather.

The earth's friction to the air and the air's viscosity are one part of the story. If we look at smaller scales, we need more details. All of these details are significant on small scales, and cause things like wind, clouds, and so on - what is called weather.

If you move towards larger scales, small details tend to get lost. For instance, it took us centuries to move from Newtonian mechanics to Quantum mechanics, simply because the small-scale details are important and it is hard to look at smaller scales. (This, sort of, is also why we can predict climate change but not weather).

### 32. Gravity doesn't exist.

Is it even necessary to look at this

*again*? Let us think of lift - the lift of wings. Let us, for a moment, consider a bird flying around in a windless environment. The frame of reference is sufficiently small that we can ignore that the earth is spinning.Sketch for how lift works |

The bird has its wings at a small angle. It starts by flapping them - it pushes down on the earth. Because of Newton's third law, pushing down on the air with your wings is the same as the air pushing on your wings. In terms of forces, that is - the bird is still providing the energy to do this. Either way, it now has some speed. By keeping the wings at a slight angle, it has to deflect the air flow.

If you're wondering what that means, think of the following (Figure to the right). You take a swimming pool - one of those roundish blow up things for the summer? You put a water hose in it. The water hose provides a strong water flow. Now, hold something - say, a plate - in front of it. Don't put it perpendicularly, but just at a small angle. Because the moving water bounces off it, a force is exerted that pushes it away.

Same principle for wings at an angle. So, the fact that it is moving and that these wings are at an angle means that the force will be partially against the direction of movement and partially upwards.

Let's look at that one more in depth, with a nice GIF. The thing about flows is that they are flowing - and they need to keep flowing. The flowing part of the picture - like the water hose we were talking about - interacts with the rest of the water via viscosity. As a result, the rest of the water or air responds. From here on, we will call it a fluid, from Latin. It's either gas or liquid, so it fits our analogies.

The interaction causes the entire picture to change. As you can see in the image below, the entire flow over the airflow is affected. In the image, we see that the flow before the wing already starts to deflect upward/downward. And, afterwards they combine again. The thing is that there is 'less' flow above and in front of the wing than there is below and in front of it; simply note the highest point of the wing in the picture. As a result, the fluid flow over the wing has to be faster than the wing, while the flow below it is usually slower.

This animation displays how the wing and its shape affect the entire flow. [Wikipedia/Lift] |

It's called Newton's Law of gravity. A

*scientific law*is, in essence, just the relation we see in the data. Nothing more or less; it doesn't contain an explanation. Einstein's theory of general relativity

*explains*what we measure, and therefore it is a

*scientific theory.*(NB: Theory also means it hasn't been falsified, despite a great many attempts).

I'm going to use this part as my explanation for lift, so we'll talk about a bit more generally. The reference is

*Introduction to Flight*. In section 5.19, he speaks of the various explanations that are put forth to explain lift, and that there is a debate centring on which particular explanation is the most fundamental.

He starts with what he himself advocates as the most fundamental explanation of lift. He writes there that lift is mainly due to the imbalance of the pressure distributions over the top and bottom surfaces of the airfoil. Why?

*``Because of the law of mass continuity - that is, the continuity equation - the flow velocity increases over the top surface of the airfoil more than it does over the bottom surface."*This is what we spoke about earlier.*``Because of the Bernoulli effect, the pressure over the top surface of the airfoil is less than hte pressure of the bottom surface."*The Bernoulli effect is that when the velocity increases, the pressure decreases.

These two reasons are, by the way, equivalent. As far as I can tell, these two are not distinct, but different formulations of the same physical mechanism. It thus follows that ``

*Owing tot he lower pressure over the top surface and the higher pressure over the bottom surface, the airfoil experiences a lift force in the upward direction."*The second explanation is the one we considered before; the deflection of the fluid flow. I completely agree with the following: ``

*However, this explanation really involves the effect of the lift, not the cause.".*

*A third explanation is called*

*the circulation theory of lift*. ``

*However, this turns out to be not so much of an explanation of lift per se, but rather more of a mathematical formulation for the calculation of lift for an airfoil of given shape".*With what he explains, the essence is to think of the flow as having two different velocities. One is the

*uniform flow*, and the other is

*pure circulation*. Together, they fit; because the circulation is clockwise, the total velocity at the bottom is the difference between uniform and pure circulation, while that at the bottom is the sum of these. As he explains it here, it is really just a mathematical description rather than a distinct explanation.

So - you now know how lift is generated, and why that doesn't contradict the theory of gravity.

### 33. Fluids aren't fluid

This really isn't that much different from the previous one. This time, it is claimed that because gravity holds the water down, fish can't swim.

Let's talk about Eureka. It means

Let's talk about Eureka. It means

*give me a towel*. It is known because the parable of the discovery of buoyancy has Archimedes sitting in a bathing pool, thinking about why things float. At some point, he jumps out, yells ``*Eureka!*'', signifying that he solved it.
The principle is really simple. Consider an object that is submerged. What, exactly, are the forces on it? Well, the force on it is simply pressure - the random bouncing of water molecules. How big is that force? Well, if the object wasn't there - if there was just water there - then the upwards force exerted by the rest of the water would be just sufficient to hold it up. Evidently it is the mass, which is the volume times average density, times the gravitational constant (that is, $g\approx9.81\:\left[\frac{N}{kg}\right]$). Suppose the object has an average density that is lower than that of water; it generates less gravitational force (pointing downward) than the upwards force (pointing upwards) of buoyancy. Surprise, it floats. Eureka!

Bony fish have a rather interesting organ called a swim bladder. Essentially, it is a bladder - a sort of balloon. Gas, mostly oxygen absorbed in the blood, can fill up the balloon. Likewise, it can also be forced out - I'm not sure of the details, but I assume it is by compressing the gas so that the gradient causes the gas to get reabsorbed.

With this, they can essentially control their average density. When it wants to go up, the density decreases so that buoyancy makes it float. When they want to go down, the density increases so that they can go down. Mind you, pressure also increases when you go down, so there's a maximum of how far they can go down.

Cartilaginous fish - like sharks - don't have a swim bladder.But in #32, we discussed lift - you won't be surprised to learn that they use their fins, in particular their pectoral fins. As far as I can see, average density varies greatly over the various species, as does fin shape - this tells you something about how versatile their system is.

Bony fish have a rather interesting organ called a swim bladder. Essentially, it is a bladder - a sort of balloon. Gas, mostly oxygen absorbed in the blood, can fill up the balloon. Likewise, it can also be forced out - I'm not sure of the details, but I assume it is by compressing the gas so that the gradient causes the gas to get reabsorbed.

With this, they can essentially control their average density. When it wants to go up, the density decreases so that buoyancy makes it float. When they want to go down, the density increases so that they can go down. Mind you, pressure also increases when you go down, so there's a maximum of how far they can go down.

Cartilaginous fish - like sharks - don't have a swim bladder.But in #32, we discussed lift - you won't be surprised to learn that they use their fins, in particular their pectoral fins. As far as I can see, average density varies greatly over the various species, as does fin shape - this tells you something about how versatile their system is.

### 34. Ships don't factor in the curvature.

A sextant. Taken from wikipedia |

So, why did they use longitude and latitude, then? These are just angles in a spherical coordinate system - so why use that instead of just cartesian coordinates?

Great Circle sailing? Do they even know what a Great Circle is? That's right, it's the shortest distance between two points on a spherical surface.

Let's talk about navigation. It's the reason we often define time in GMT - Greenwich Mean Time. To the right, we see an instrument called a sextant. It is used to determine the longitude/latitude of the navigator of a ship. This is also a story of why things like Kepler's Laws and Newton's Mechanics were so very interesting - back then, we didn't have GPS and this sort of navigation was

Let's talk about navigation. It's the reason we often define time in GMT - Greenwich Mean Time. To the right, we see an instrument called a sextant. It is used to determine the longitude/latitude of the navigator of a ship. This is also a story of why things like Kepler's Laws and Newton's Mechanics were so very interesting - back then, we didn't have GPS and this sort of navigation was

*extremely important*.
The Oxford Museum of Science has a very large collection of sextants in it. Yes, It was fun to look at those, albeit the museum isn't exactly a

*museum*. A collection is a better word. Anyway, I got my hands on a sextant in an old shop, but it was too expensive and we didn't have surplus space, since we went by plane.
Essentially, you need a number of things. First, a nautical almanac - the first one was published by the HM Nautical Almanac office on the site of the Royal Greenwich Observatory in 1767. Second, a sextant - a precision tool. Third, a chronometer - basically a really good clock - synchronised to the reference point of the Almanac. Fourth, accurate charts. Fifth, a mathematical tool.

The nautical almanac essentially contains tables that tell you the hour and the position on the Earth's surface at which a specific celestial body (the sun, the moon, several planets, etc.) are directly overhead. We will, for simplicity, assume that we are using a nautical almanac, with as reference point the Royal Greenwich Observatory (RGO).

The sextant allows you to measure the angle between a celestial object and the horizon.The clock tells you what time is at the RGO.

From the time at the RGO, you can determine the celestial coordinates of the object when seen from the RGO. You use the mathematical tool/formula to use your data, so the celestial measurement and the reference data from the RGO to determine your longitude.

If you're on the Northern Hemisphere, you can easily find your latitude from the angle between the horizon and polaris. This is extremely convenient - and the entire reason why polaris is so important. On the Southern Hemisphere, you use the polaris australis or more properly sigma octantis for the same purpose.

If you're on the Northern Hemisphere, you can easily find your latitude from the angle between the horizon and polaris. This is extremely convenient - and the entire reason why polaris is so important. On the Southern Hemisphere, you use the polaris australis or more properly sigma octantis for the same purpose.

That gives you your longitude and latitude. You are now know where you are, and can grab your most applicable navigational chart. This means a chart of a rather small local area, so that the mercator projection is largely applicable.

The image shown in the article is from Brother Soft - the modern, electronic version of such charts. Because you don't want to be stranded on sea without a proper chart - say, because the electronics fail - modern seamen still have the good old hardcopies around.

*celestial navigation*.

It is important to note that it is the angle of the horizon with the celestial object that is important. The celestial object is far enough away that the same results couldn't be reproduced on a flat surface - the fact that celestial navigation works is testimony to the earth being an oblate spheroid.

### 35. A random assertion.

This is just an assertion.

I'm going to make a few guesses here:

*``The fact that many captains navigating south of the equator assuming the globular theory have found themselves drastically out of reckoning, more so the farther south travelled, testifies to the fact that the Earth is not a ball''.*I'm going to make a few guesses here:

- The mathematical tools available or used were only valid for the northern hemisphere, so using them when they don't apply gets you lost.
- Trying to navigate by polaris - which isn't visible - went wrong because the star they were using wasn't actually polaris.
- The so-called fact is just a lie.

### 36. Something about some Captain being perplexed.

I'm going to make a guess here, too. That's what happens when something is asserted without reference.

I'm going to assume that, being a Captain, he knew how to use celestial navigation. If I'm not mistaken, it was properly the job of his navigator but I imagine that the captain would be able to do the job if required.

I'm also going to guess he wasn't a physicist. He'd been taught the earth was roughly spherical. However, as Newton already predicted, it's not a sphere - it is an oblate spheroid. It's flatter at the poles. They were charting, mostly, so I'm not entirely sure why it is claimed they found themselves outside their reckoning.

If it's their dead reckoning - basically, saying I travelled this far in that direction - then I'm not surprised that they found the wrong result assuming a completely spherical earth - because, you know, it's not completely accurate and they were at the most inaccurate point. We'll talk about it a bit more, soon.

### 37. Another one.

It is claimed that Lieutenant Charles Wilkes had the same problem, sometimes being over 20 miles in less than 18 hours.

I'm pretty sure that it's not accounting for the flattening in the dead reckoning. So the dead reckoning is the following. First, you take your previous position; say, $(\theta, \phi)$. I'm using $\theta$ as the

*polar*angle and $\phi$ for the

*azimuthal*angle [Wikipedia/spherical coordinates/].

I'll write down some mathematics first, and then explain what it means.

$$\frac{d}{dt} \underline{r}\left(r,\theta,\phi\right)

=\frac{\partial \underline{r}}{\partial r}\frac{dr}{dt}+ \frac{\partial \underline{r}}{\partial \theta}\frac{d\theta}{dt}+ \frac{\partial \underline{r}}{\partial \phi}\frac{d\phi}{dt}$$

This is just something called the full derivative. You assume that the coordinate is a function of the distance from the centre, the polar angle and the azimuthal angle. Then, the time derivative - also known as velocity - is simply the partial derivative in one coordinate times the rate of change of that coordinate, plus the same of the other ones. For spherical coordinates, we get something called the line element, which is basically the velocity times an infinitesimal small time. It's the instantaneous displacement, given by:

$$ d\underline{r} = dr \:\hat{r} + r \left(d\theta \:\hat{\theta} + \sin{\theta }d\phi\:\hat{\phi}\right)$$

What this means is that the displacement is the sum of:

- The change in radial distance, the distance from the centre, in the direction of the radial distance.
- The radial distance times the change in polar angle in the direction of the polar angle.
- The radial distance times the change in azimuthal angle, times the radial distance, times the sine of the polar angle. The sine of the polar angle accounts for the shrinking circumference (seen from the south-north pole line or rotational axis).

So they assumed that the radial distance didn't change. It did, because of the flattening of the earth's poles. They assumed it was constant when it was not; as a result, they overestimated the change in longitude and latitude.

I think this is a likely explanation; without a proper reference, I can't see if it is the actual explanation.

### 38. Let's not give a reference #MXXXIII.

To quote: ``

*This misfortune happened to a fine frigate, the Challenger, in 1845*.'' I have no idea who `Reverend Thomas Milner' is, and wikipedia doesn't know.
Eight ships of the Royal Navy of Britain have been called HMS Challenger. However, none was named thusly in 1845. The last one before it got wrecked in 1835, off Chile. The next one after was built in 1858. (From wikipedia and books.google "

*Ships of the Royal Navy*").
``

*How came Her Majesty’s Ship ‘Conqueror,’ to be lost?''.*The only ship that could apply was the HMS Conqueror (1855), a 101-gun first rate which was launched in 1855 and wrecked in 1861. It was wrecked off Rum Cay, which is an island of the Bahamas. Also known as the Northern Hemisphere. It was due to a navigation error, apparently. [Wikipedia/HMS conqueror 1855)In conclusion, it appears that the named events in this anecdote didn't exist.

### 39. How to fail at calculations.

They state here that the straight line distance between Sydney and Nelson is $1550$ statute miles ($2494.483$ km). And, the longitude difference is 22 degrees 2'14'' ($22.0372$ degrees).

They say that the entirety would measure $25182$ miles ($40526$ km). So, that gives a radius of $6449$ km, versus the $6371$ km average radius.

To calculate the distance, we must first convert to radians. We find that $22.0372\cdot \frac{\pi}{180}\approx 0.3846$. If we assume the average radius is okay ($6371$ km), we find a distance $ds = r d\phi = 0.3846 * 6371\ = 2450.40:\left[\text{km}\right]$ - that's pretty nice.

In fact, Nelson lies to the north of Sydney:

- Sydney at 33.8650° S, 151.2094° E [Google/Sydney Longitude Latitude]
- Nelson at 41°18' S, 173°16' E [MapsOfWorld/Latitude Longitude / New Zealand]

This gives a difference of 7.435°, 22.0573° - so the longitude difference agrees. We just calculate the latitude difference too, assume the radius is constant (the ~$7$ km difference won't matter much) and use the Pythagorean Theorem to calculate the distance. The result is $2588$ km. Again, not too bad.

So our result is $2588$ km assuming a perfect sphere (it isn't) versus the distance as they tell it being $2494$ km. Of course, we can't check the longitude/latitude they took for the $1550$ statute miles. Additionally, we are assuming a straight line instead of the correct

*Great Circle*(shortest difference over a spherical surface) - it's not weird that we don't get the exact result.

I'm going to say that these calculations match the given straight line distance.

### 40. Apparently, latitude doesn't matter.

The two points named are:

- Cape Horn, Chile, Latitude:-55.983333° Longitude:-67.266667°
- Melbourne, Australia, Latitude:-37.814251°Longitude:144.963169°

They then state a distance between these two points and calculate the radius from the longitude difference. I'm sorry, but the latitude matters too. Using the Pythagorean approximation we used earlier, the distance they give, a radius of $6371$ km and the difference in longitude they give, we find that

$$d\theta\approx\sqrt{\frac{s^2}{r^2}-d\phi^2}\approx 15.5272^o$$

Drat, it was ~$56-38=18$. Well, that's the spherical earth disproven, that is.

Oh, wait. They still agree.

### 41. Let's do the same thing wrong, again.

Are you in the mood to check their calculations again? I'm not. It seems pointless.

Let's do one check, though. They state that ``

*According to the ball-Earth theory, the circumference of the Earth at 37.5 degrees Southern Latitude should be only 19757 statute miles, almost six thousand miles less than such practical measurements.''*

Geocentric radius of rather easy to calculate. It is a function of geodetic latitude. I put it in matlab with the $37.5^o$, just to check their claim of it being 19757 statute miles ($31796$ km). The radius calculated was $6370.3$ km and the circumference was $40025$ km, or 25000 miles. For $35.5^o$ south, it was $6371$ km radius and $40030$ km circumference. That's the 25000 miles that they are yelling about it isn't.

I'm confused - is it so hard to find these formula on Wikipedia?

I'm confused - is it so hard to find these formula on Wikipedia?

### 42. Backing up your assertions is a lie.

Something about Antarctic circumnavigation being far too long.

Captain James Cook circumnavigated Antarctica. [National Science Foundation] |

Most races or solo attempts start from Europe. Due to the configuration of the continents, sailing around the world consists in sailing around the Antarctica continent, passing south of Cape Horn, Cape of Good Hope and Cape Leeuwin. [Wikipedia/World sailing record]

The actual coastline length is $11165$ mi ($17968$ km) [Wikipedia/Antarctica]. And, of course, the coastline isn't equal to the circumference of the earth there. Antarctica isn't shaped like a disc. You can easily see this from the given surface area; if it was disc-shaped, you'd be able to make a good estimate at the circumference by calculating the apparent radius. The area is $A = \pi r^2$ and the circumference is $ 2 \pi r$. Some elementary mathematics give you that the circumference of disc-Antarctica is $\sqrt{4\pi A} \approx 13260 \neq 17968$ km.

So what's the issue here? Well, neither Ross nor Cook exactly took a straight route. I mean, look at the image to the right.

And here's the flat-earthers pretending he went in a nice circle. The key here is that 'circumnavigated' means 'navigated around something'. It doesn't mean 'went in a circle'. The word 'circumference' is apparently used for two reasons; one is because the expression is 'to go around', even if it is a square object. The other is because perimeter-navigated doesn't sound as nice.

Because

So what's the issue here? Well, neither Ross nor Cook exactly took a straight route. I mean, look at the image to the right.

And here's the flat-earthers pretending he went in a nice circle. The key here is that 'circumnavigated' means 'navigated around something'. It doesn't mean 'went in a circle'. The word 'circumference' is apparently used for two reasons; one is because the expression is 'to go around', even if it is a square object. The other is because perimeter-navigated doesn't sound as nice.

Because

*circum*comes from*circumference*, which is the perimeter of a circle, people will naturally think of circles - even if there's absolutely no circle present.### 43. Cold air is a lie, too.

First, let's consider Great Circles. We've already mentioned them a few times, but we'll go over them now in some depth.

**Warning: Some elementary calculus is ahead - skip to 43b if you don't want to see calculus.**If you haven't covered this in your education, that's okay. I'll try to explain as much as seems necessary. I*will*assume that you know what derivatives and integrals are.
Let us, for simplicity, assume the earth is a perfect sphere. The calculation can be done without this assumption, but it becomes more complex. The question we must ask ourselves is what the distance travelled on a sphere is?

We'll look at the time derivative for a function $\underline{r}$, which is commonly known as the velocity $\dot{\underline{r}}\equiv\underline{v}$, where we will follow Wikipedia and use $\theta$ as the inclination from the polar axis:

$$\begin{align*}

\dot{\underline{r}}&=\frac{d\underline{r}}{dt} \\

&=\hat{r}\dot{r} + \hat{\theta} r\dot{\theta} + \hat{\phi} r \sin{(\theta)}\dot{\phi}

\end{align*}$$

Assuming the perfect sphere, $\dot{r}=0$. As a result, the absolute displacement becomes:

$$\begin{align*}

\dot{\underline{r}}&=\frac{d\underline{r}}{dt} \\

&=\hat{r}\dot{r} + \hat{\theta} r\dot{\theta} + \hat{\phi} r \sin{(\theta)}\dot{\phi}

\end{align*}$$

Assuming the perfect sphere, $\dot{r}=0$. As a result, the absolute displacement becomes:

$$\begin{align*}
\left|\dot{\underline{r}}\right| &= r \sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}}
\end{align*}$$

Mind you, here we've taken $\underline{r}(t=0)$ to coincide with the north pole. We'll talk about how that assumption complicates things in a minute.

Thus, for the displacement we find:

$$\begin{align*}

S &= \int_{t_0}^{t_1} dt \left[ r \sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}} \right]

&= \int_{t_0}^{t_1} dt F\left(\theta, \dot{\theta}, \dot{\phi}\right)

\end{align*}

$$

$ F\left(\theta, \dot{\theta}, \dot{\phi}\right)$ is just a renaming; I wanted to do this bit somewhat more generally. It's also the key to the Lagrange and Hamilton formalisms of classical mechanics. The idea is to go from the point $(r, \theta(t_0), \phi(t_0))$ to $(r, \theta(t_1), \phi(t_1))$. If you want a reference, it's Numerical Methods in Scientific Computing.

Let us consider a small parameter $\epsilon_i$ and an arbitrary function $\eta_i(t)$ such that $\eta_i(t_0)=\eta_i(t_1)=0$. Note the subscript; $\epsilon_1$ does not need to be $\epsilon_2$! Let us call the solution to our problem the set of functions $\left\{ \hat{\theta}(t), \hat{\phi}(t)\right\}$. The braces indicate that these two go together. Then we can write an arbitrary solution as $\left\{ \hat{\theta}(t) + \epsilon_1 \eta_1(t), \hat{\phi}(t) + \epsilon_2 \eta_2(t)\right\}$. However, we want to have the best function - that function which is closest to the real solution. So, we minimise with respect to $\epsilon_i$. As a result, we find:

$$\begin{align*} \frac{d}{d\epsilon_1} S &= \int_{t_0}^{t_1} dt \left[ \frac{\partial F}{\partial \theta} + \frac{\partial F}{\partial \dot{\theta}} \dot{\eta_1}\right] \\ \frac{d}{d\epsilon_2} S &= \int_{t_0}^{t_1} dt \left[ \frac{\partial F}{\partial \dot{\phi}} \eta_2\right]\end{align*}$$

Now, we'll look at the first one - it is the most interesting. If you recall Integration by Parts, it's easily solvable. Remember the product rule for differentiation? $d(uv) = v du + u dv$. Well, integrate that and you'll find $ \int u dv = uv - \int v du$, which is the point of Integration by Parts. We'll do that:

$$\begin{align*} \frac{d}{d\epsilon_1} S &= \int_{t_0}^{t_1} dt \left[ \frac{\partial F}{\partial \theta} - \frac{d}{dt}\frac{\partial F}{\partial \dot{\theta}}\right]\eta_1 + \left[\eta_1(t)\frac{\partial F}{\partial \dot{\theta}}\right]_{t_1}^{t_2} \end{align*}$$

But, we said that $\eta_i(t_0)=\eta_i(t_1)=0$. So the last term is gone.

Since we considered arbitrary $\eta_i (t)$, the requirement here is that the term in brackets equals zero. This requirement is called the Lemma of Dubois-reymond. Either way, the result is something called the Euler-Lagrange Equations (for this particular $S$):

$$\begin{align*} \frac{\partial F}{\partial \dot{\phi}} &= 0 \\ \frac{\partial F}{\partial \theta} - \frac{d}{dt}\frac{\partial F}{\partial \dot{\theta}} &= 0 \end{align*}$$

So, we are now ready to fill in these equations for our $F = r \sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}} $. We will find a curious result after considering only the first Euler-Lagrange equation.

$$\begin{align*} \frac{\partial F}{\partial \dot{\phi}} &= 0 \\ \frac{r}{2\sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}}} \sin{^2\theta} \cdot 2 \dot{\phi}&= 0 \\ \dot{\phi}&=0 \\ \phi &= \text{constant} \end{align*}$$

That's right; $\phi$ is a constant. When we substitute that, we find $F = r \dot{\theta}$ and the form of $\theta(t)$ becomes immaterial.

This is a plane of constant $\phi$, through the origin of the sphere. Now, that's sufficient information to determine the great circle path. If you have two points, you determine the plane through the two points and the origin. The easiest way to do this is use the two points to get the orthonormal basis of the plane. So, first Gram-Schmidt:

$$\begin{align} \underline{e_1} &= \underline{r}_1 \\ \underline{e_2} &= \underline{r}_2 - \underline{r}_1 \frac{ \underline{r}_1\cdot\underline{r}_2}{\underline{r_1}\cdot\underline{r_1}}\end{align}$$

The great circle is defined by lying in the plane and having distance $R$ to the origin. That's a rather simple equation for the parametrised curve:

$$\begin{align} \underline{r}_\text{great circle}(s) &= R \left( \underline{e}_1 \cos{s} + \underline{e}_2 \sin{s}\right)\end{align}$$

Now, just choose $s_1$ such that the great circle is at $r_1$ at $s_1$ and $s_2$ such that the great circle is at $r_2$ at $s_2$, and you have the great circle path.

Contrary to the Image fabricated by the flat earthers, I didn't really fabricate anything. What you see here is comparable to a satellite image - it's just a floating camera above the south pole.

Now, we can also ask google questions such as '

The logical way to fly from Sydney to Santiago would still be a great circle. So no, it wouldn't be a straight line. Here's the route on a Mercator Projection. If you don't know what map projections are - and I should've raised this point before - now is the time to learn. I'll refer you to a neat webcomic to get you started.In the meantime, here's the Great Circle route from Sydney to Santiago (Chile).

I'm not sure about their detours - not only is there apparently a direct route now, there's always been flights with a small stop in New Zealand [traveller.com]:

It is for that reason that airplanes regularly overbook by about 10%. It's about the amount of passengers that ultimately don't show. Sometimes, this means they will have some passengers that can't board. These will get a compensation. This happened to me on my flight from Amsterdam to Geneva, for L'ecole des Physique des Houches. I missed one single introduction lecture, but got free lunch, some money (which ultimately was spend towards books, boxes for the tv and a new tablet).

So, what do they spend the money on? Well, first look at the infographic on the left. I hadn't expected to find such a conceptually simple visual overview, but I did - and extremely fast, too. It was the third result on google?

As for the question of how non-stop flights are chosen, we'll turn to that in #46.

Well, some airports are so-called hubs. The idea is not uncommon and is, by far, not limited to avionics:

International routes, such as Cape Town - Buenos Aires, are flights that require a long-range plane (expensive), and thus require many passengers (unlikely). However, the same isn't true for a hub. Flights to hubs are far more feasible because there are extremely large numbers of other flight routes (hundreds) that connect, and as a result the flight is more feasible.

So a small location, in terms of flight routes, would connect to a hub. Then, you travel to another hub. And then you finally come to your destination. How to determine hubs? Wikipedia has an (incomplete?) list.

For instance, from Cape Town (hub), you travel to Johannesburg (hub), to São Paulo (hub), to Buenos Aires (hub). (Image to the right)

Seems like that makes sense, doesn't it? After all, when you want to travel between two large cities in your country, you don't use the shortest route in terms of route. No, you'd rather use the high way - you can go much faster on a highway. And because highways run past hub cities, you'll take a bit of a detour, but only in terms of distance - you still save time. The analogy is sound, but for airplanes it is costs that are saved, not time.

There is but one good response, and I put it into a picture. If you're wondering, that's a fictional character from Stargate: Atlantis, a very intelligent, sceptical scientist. I liked the picture and I'm bad at memes, so this'll have to do.

The factors that can break the symmetry are probably the axial tilt and refraction of sunlight.

We'll consider the phenomenon a bit more in #54.

Yes, their model completely ignores all evidence to the contrary. That's sort of the point of my article.

Their assertions depend, as usual, on not showing that anything is actually true. In particular, it is noteworthy that they don't define the Horizon on their flat earth model.

As you can see, it is barely visible in the best of cases. At heights below the $4000$ m ($13000$ ft), I wouldn't bet on seeing it. The article in Applied Optics also describes some other obscuring effects. So, what'd it look like? Let me pull from my own images (for once). I took it out of an airplane flying from Amsterdam to Geneva, at a height of 9.2km ($30000$ ft). As you can see, it is barely visible. As the Applied Optics image outlines, there is a slight skewing because my camera isn't well aligned to the horizon. I could see it, just barely, but I am also wearer of glasses. So I can't entirely trust that observation. Indeed, comparison to Fig 5. of the Applied Optics paper shows us that my camera is affecting the results, and likely my glasses are as well.

As a final remark, someone from my study had something cool quite recently. He's on the TU Delft DARE project, and they launched the Stratos II+ rocket. It got to quite a high altitude. But what's relevant here is that it is a rocket with mounted cameras on it - so you can see around it when it is ascending. For that purpose, I link to the youtube video, slowed, here. Youtube slow is a website I've used before; it allows you to watch it slowly. The rocket is spinning quite rapidly, which should help it stabilise. Note that the left lens does seem to be a non-fisheye one, but I can't say for sure.

In conclusion, apparent flatness isn't equal to actual flatness. There are better methods available to show it. We've known since

If you're wondering, that ends up in $5.6035 \cdot 10^{-4} \sqrt{h}$. So, standing on the top of the Big Ben (96 m), you would be able to resolve.. $0.0055$ radians or $0.3146^o$. We've already discussed the angular resolution of the eye, which is $1^o$. You would be able to resolve such a thing when you're at about a kilometre high.

The tallest building in the world is the Burj Khalifa in Dubai, but it is only $828$ m high. Can you see it? I think I can, but I'm not sure.

I'll point to the scientific american article again, and include this quote:

You see what I mean? It's also telling that these are always low-quality videos. The original video was actually evidence

As for angles, we already discussed these. You probably won't be able to see this until you're at a height of about a kilometre.

Here, they claim N.D. Rowbotham "

No, what it mostly did was this:

Mirages are optical phenomenon due to the different densities of air causing different bendings of light [Wikipedia/Mirage]. A well known mirage effect is that the sun appears flattened when setting. They are the better known one, but this sort of thing happens all the time. Looming is fun too [Wikipedia/Looming]. I first saw that in the plane to London, when I saw a lot of boats forming a sky armada. Rather than believing my eyes and believing the Empire was about to strike, I thought about refraction for a bit and continued to read my book. The green flash is good evidence, too.

I won't redo the calculation, but it has been done. The estimate for the horizon distance becomes $3.86\sqrt{h}$, due to light bending. There's a lot of assumptions going into that, largely because refraction depends on atmospheric results - basically, the weather.

I don't know the particular atmospheric conditions that are present on '

Denial of refraction is a hard one. Above, I've already pointed to a few points of evidence. Other things that depend on it are telescopes, glasses, microscopes and cameras. Probably a lot more, too. The underlying theory of light is so very, very settled by now that I can't imagine someone denying it. Of course, even if I can't imagine it, I know someone will, anyway. The underlying theory is the wave theory of light, from which Snell's law can be derived. And below that lie Maxwell's equations. Which are consistent with quantum mechanics, and on which a lot of work is based.

It doesn't really matter. The height of Bear Mountain allows for a horizon at $76$ km. However, what we forgot to include is that New York City is at 500 feet elevation itself.

Let's use my intuitive guess from earlier and just add those. That gives us $89.9851$ km, which still isn't enough.

So yes, part of the

A quick calculation using the height from Bear mountain, the 500 feet elevation of new york, and inversing the horizon distance formula, tells me that any building higher than $124.2365$ m is visible. That leaves

Miraggi accadono, faccia o requiescat in pace. I don't speak Italian, but google translate does. And I put in Ezio Auditore da Firenze's `requiescat in pace', because references.

Both have extended range because they use a Fresnel Lens, apparently.

The rest of the difference can probably be explained by a mirage.

You do realise airplanes, rockets, satellites and various other things are built using Newton's law of gravity? Good.

So here's the Cavendish experiment:

Note that the Cavendish experiment has been reproduced a great number of times.

I went to google, and I did something

You don't? Why not? Because your

You are worthy of endless ridicule if you keep an idea when it is so very, very clearly refuted by common sense, daily experience, high school level physics (and beyond), and so on. You literally have the option of looking at a photograph of the Earth. But no, an elaborate conspiracy involving basically everyone that doesn't believe you has to be present because you want to

Show us these, instead of showing us silly pictures. Explain, predict, and calculate, instead of asserting and moving on to the next bit of your gish gallop.

Let's think about a sun that is close by, and a layer of clouds with holes in it. I've drawn a sketch to the right.

As you can see, a lot of rays won't even make it to the ground. This is because the sun, being so close, doesn't have rays parallel to the 'tunnels' through the clouds.

Now, what would it look like with a far away sun? Again, the sketch is to the right. You see that because of the distance from the sun, it has a lot of rays that are sufficiently parallel to the tunnels to go through. At ground level, the rays are quite a bit away from each other. The final question is: What would happen if you add perspective?

The curvature is very flat, compared to what most people would think. Still, it does affect the angle of incidence and it is the reason that you have this long stretch of reflection.

A flat earth with a sun that is close, and especially the sea as the surface, would act more like a mirror. While I won't tell you to grab a mirror and refract the sun in it, you can use a dim light and do the same. You don't get this line, but a reflection spot and that's it.

One thing that comes to mind is that solid stone is solid. You can't accidentally move it, and it can't be moved (or not easily) by natural causes, such as (weak) earthquakes.

Most amateur telescopes are mounted on tripods, and we put the tripods on a hard surface. Not because of 'not moving a hair's breadth' - which is important with a very limited view angle - but because it needs a solid base. If you put in damp soil, for instance, then it will slowly start sinking into it. The effect is more pronounced when not all three of the legs are in equally damp soil. Note that a telescope usually has a counter weight, so that even a relatively small one like me has total mass of 6-7 kg. (The counter weight is about 5).

However, one has to appreciate that stars are bloody far off.

Which means you need extremely accurate readings - far more accurate than those by human eyes, which are rather bad. From Wikipedia, we get:

The

A lot of apparently intentional vague claims here. I'm going to ask a simple question here, and then explain the answer. Why is the sun's light "golden" and the moon's light "silver"?

To the right, we see an image of the moon. Note how it is not blueish.

There is a

For low illumination levels, the different behaviour is that it perceives blue far stronger than the other colours. Normally, it is a reasonable measurement of the spectrum of the light (how strong each monochromatic colour is). When illumination is low, it is not; it suppresses everything but blue. And that is why moon light is blueish/silver; it's white with slightly more blue than you normally perceive.

For high illumination levels, the different behaviour is that it perceives red stronger than the others. As a result, in daylight things seem more reddish, more golden.

There's an interesting figure called the Kruithof Curve. It's a bit complicated, so if you look into it you will also need colour temperature. It basically shows you colours on one axis, illumination on the other and perceived colour as an intensity-plot.

I'll directly address those. First, moonlight - even when concentrated over 6000 times - didn't heat anything. I'm not surprised. You shouldn't be either.

The sun is at some $1000 \left[\frac{w}{m^2}\right]$. So, moonlight is at a thousand of that. A lens of 30 inches in diameter was used. We can calculate the surface area; the diameter is $76.2$ cm, the radius is $38.1$ cm, and the surface area is $\pi r^2 = \pi 0.381^2 \approx 0.456 \: [m^2]$. Therefore, the heating done by this concentrated moonlight is about $0.456 \cdot 6000 \cdot 0.001 \approx 2.74 W$.

Suppose that, for no apparent reason, the heated surface is a thermometer stuck in a glass of water (0.2 L, standard volume). How much would this water heat per second? Well, 0.2L of water is about 0.2kg of water, and it has heat capacity 4185.5 Joule per Kg per Kelvin. Therefore, it needs 837 J for one degree of heating. Using the concentrated lunar light above, that's about half a minute. This is very slow heating, and the heat would probably dissipate at a similar rate.

As a physicist, I'm going to state this clearly. Moonlight is just light; there's nothing special. It's made of photons, it broadly obeys the black body spectrum, and so on. There's nothing fancy.

Mind you, here we've taken $\underline{r}(t=0)$ to coincide with the north pole. We'll talk about how that assumption complicates things in a minute.

Thus, for the displacement we find:

$$\begin{align*}

S &= \int_{t_0}^{t_1} dt \left[ r \sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}} \right]

&= \int_{t_0}^{t_1} dt F\left(\theta, \dot{\theta}, \dot{\phi}\right)

\end{align*}

$$

$ F\left(\theta, \dot{\theta}, \dot{\phi}\right)$ is just a renaming; I wanted to do this bit somewhat more generally. It's also the key to the Lagrange and Hamilton formalisms of classical mechanics. The idea is to go from the point $(r, \theta(t_0), \phi(t_0))$ to $(r, \theta(t_1), \phi(t_1))$. If you want a reference, it's Numerical Methods in Scientific Computing.

Let us consider a small parameter $\epsilon_i$ and an arbitrary function $\eta_i(t)$ such that $\eta_i(t_0)=\eta_i(t_1)=0$. Note the subscript; $\epsilon_1$ does not need to be $\epsilon_2$! Let us call the solution to our problem the set of functions $\left\{ \hat{\theta}(t), \hat{\phi}(t)\right\}$. The braces indicate that these two go together. Then we can write an arbitrary solution as $\left\{ \hat{\theta}(t) + \epsilon_1 \eta_1(t), \hat{\phi}(t) + \epsilon_2 \eta_2(t)\right\}$. However, we want to have the best function - that function which is closest to the real solution. So, we minimise with respect to $\epsilon_i$. As a result, we find:

$$\begin{align*} \frac{d}{d\epsilon_1} S &= \int_{t_0}^{t_1} dt \left[ \frac{\partial F}{\partial \theta} + \frac{\partial F}{\partial \dot{\theta}} \dot{\eta_1}\right] \\ \frac{d}{d\epsilon_2} S &= \int_{t_0}^{t_1} dt \left[ \frac{\partial F}{\partial \dot{\phi}} \eta_2\right]\end{align*}$$

Now, we'll look at the first one - it is the most interesting. If you recall Integration by Parts, it's easily solvable. Remember the product rule for differentiation? $d(uv) = v du + u dv$. Well, integrate that and you'll find $ \int u dv = uv - \int v du$, which is the point of Integration by Parts. We'll do that:

$$\begin{align*} \frac{d}{d\epsilon_1} S &= \int_{t_0}^{t_1} dt \left[ \frac{\partial F}{\partial \theta} - \frac{d}{dt}\frac{\partial F}{\partial \dot{\theta}}\right]\eta_1 + \left[\eta_1(t)\frac{\partial F}{\partial \dot{\theta}}\right]_{t_1}^{t_2} \end{align*}$$

But, we said that $\eta_i(t_0)=\eta_i(t_1)=0$. So the last term is gone.

Since we considered arbitrary $\eta_i (t)$, the requirement here is that the term in brackets equals zero. This requirement is called the Lemma of Dubois-reymond. Either way, the result is something called the Euler-Lagrange Equations (for this particular $S$):

$$\begin{align*} \frac{\partial F}{\partial \dot{\phi}} &= 0 \\ \frac{\partial F}{\partial \theta} - \frac{d}{dt}\frac{\partial F}{\partial \dot{\theta}} &= 0 \end{align*}$$

So, we are now ready to fill in these equations for our $F = r \sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}} $. We will find a curious result after considering only the first Euler-Lagrange equation.

$$\begin{align*} \frac{\partial F}{\partial \dot{\phi}} &= 0 \\ \frac{r}{2\sqrt{ \dot{\theta}^2 + \dot{\phi}^2 \sin{^2\theta}}} \sin{^2\theta} \cdot 2 \dot{\phi}&= 0 \\ \dot{\phi}&=0 \\ \phi &= \text{constant} \end{align*}$$

That's right; $\phi$ is a constant. When we substitute that, we find $F = r \dot{\theta}$ and the form of $\theta(t)$ becomes immaterial.

This is a plane of constant $\phi$, through the origin of the sphere. Now, that's sufficient information to determine the great circle path. If you have two points, you determine the plane through the two points and the origin. The easiest way to do this is use the two points to get the orthonormal basis of the plane. So, first Gram-Schmidt:

$$\begin{align} \underline{e_1} &= \underline{r}_1 \\ \underline{e_2} &= \underline{r}_2 - \underline{r}_1 \frac{ \underline{r}_1\cdot\underline{r}_2}{\underline{r_1}\cdot\underline{r_1}}\end{align}$$

The great circle is defined by lying in the plane and having distance $R$ to the origin. That's a rather simple equation for the parametrised curve:

$$\begin{align} \underline{r}_\text{great circle}(s) &= R \left( \underline{e}_1 \cos{s} + \underline{e}_2 \sin{s}\right)\end{align}$$

Now, just choose $s_1$ such that the great circle is at $r_1$ at $s_1$ and $s_2$ such that the great circle is at $r_2$ at $s_2$, and you have the great circle path.

### 43b. No more mathematics

This is a great circle; it is literally a great circle, a circle of constant azimuthal $\phi$. Because we want to make the point, we have to consider how these would look on the really weird projection they took. I'm going to use the map they gave and use three locations. I choose:- Sydney, Australia. (
*33.9 S, 151.2 E)* - Cape Town, South Africa (33.9 S, 18.4 E)
- Port Howard, Falkland Islands, British Overseas Territories. (
*51.6 S, 59.5W*)

I made you this plot. See text. |

I wrote a Matlab script to help us. I'll briefly describe what it does. First, there is a function that, given two points, calculates the Great Circle through those two points. Second, there is a file that plots the sphere, the north pole, south pole, the three given cities whose coordinates I took off google and finally plots these.

Contrary to the Image fabricated by the flat earthers, I didn't really fabricate anything. What you see here is comparable to a satellite image - it's just a floating camera above the south pole.

Now, we can also ask google questions such as '

*distance port howard to sydney'*. Let's do that:- Port Howard to Sydney, $10 042$ km
- Sydney to Cape Town, $11 005$ km
- Cape town to Port Howard, $6 331$ km

In their image, the distances looked very similar. I don't know why; let them explain that.Anyway, I didn't put Antarctica into my image. I'm not going to, either - I wanted to demonstrate what Great Circles are. The most important thing to note is that they are

*clearly not**the straight line between two points*.
So, why do planes not fly over Antarctica? There's a number of reasons [[Wikipedia/Polar Route]:

- Most cities simply don't have a Great Circle that lies over Antarctica.
- Airlines don't schedule those flights - for instance, because nobody wants to go from Wellington to Cape Town.
- Winds; the great circle isn't the full story. In terms of fuel and time, the shortest distance might be one that takes you out of the winds.
- Furthermore, there are a few operational considerations. For instance, extended flight over those regions might cool down the fuel too much.
- Range of the airplanes. For instance, the Airbus A380 has a range of $15 700$ km and could easily make it. A Boeing 747-400 has range $13 450$ km and could make it as well. However, for a place like the Falklands these are simply too much. I see that the Falklands will soon fly with LAN airlines to Saint Helena, and half their fleet is Airbus A320 family. Which means that the range is $3 100$ - $12 000$ km. You'll note that, when we incorporate waiting times before landing, turbulence, weather conditions and so forth, this is only barely enough to make the flight from Port Howard to Sydney.

### 44. Given that we were just wrong, why don't they do things like they do?

Via Great Circle Mapper. |

I'm not sure about their detours - not only is there apparently a direct route now, there's always been flights with a small stop in New Zealand [traveller.com]:

So they don't actually make that detour. If you look at their image to the left, you see how much of a `detour' they would make on a flat earth map.LAN operates daily one-stop flights between Sydney and Santiago via Auckland, providing onward connections to over 115 destinations in South America.

### 45. A similar one, but with different cities.

Via Great Circle Mapper. |

So, now it is about Johannesburg to Perth. Let's look at the great circle route, to the right.

Do they fly it? Looks like it [South Africa.to]. Even googling the direct flight path, adding in zero stops, gives you a clear result.

I think we should stress here that airlines fly whichever route is financially feasible. For airplanes, this generally means that the plane has to be filled.

via [Wall Street Journal] |

So, what do they spend the money on? Well, first look at the infographic on the left. I hadn't expected to find such a conceptually simple visual overview, but I did - and extremely fast, too. It was the third result on google?

As for the question of how non-stop flights are chosen, we'll turn to that in #46.

### 46. Aren't you tired of this yet?

Via Great Circle Mapper. |

Like last time. First, the great circle flight path, to the right.

And then we turn to google, which tells us that there's no non-stop flights. Which, in turn, tells us that there is no market.

However, there's a fair few with multiple stops. So what's up with that?Well, some airports are so-called hubs. The idea is not uncommon and is, by far, not limited to avionics:

Local routes are, largely, more feasible. There's less costs due distances, you can use smaller (and older) planes and there are likely more people for such a flight.Airline hubs are airports that an airline uses as a transfer point to get passengers to their intended destination. It is part of a hub and spoke model, as opposed to the Point to Point model, where travelers moving between airports not served by direct flights change planes en route to their destinations.

International routes, such as Cape Town - Buenos Aires, are flights that require a long-range plane (expensive), and thus require many passengers (unlikely). However, the same isn't true for a hub. Flights to hubs are far more feasible because there are extremely large numbers of other flight routes (hundreds) that connect, and as a result the flight is more feasible.

via Google Flights. |

So a small location, in terms of flight routes, would connect to a hub. Then, you travel to another hub. And then you finally come to your destination. How to determine hubs? Wikipedia has an (incomplete?) list.

For instance, from Cape Town (hub), you travel to Johannesburg (hub), to São Paulo (hub), to Buenos Aires (hub). (Image to the right)

Seems like that makes sense, doesn't it? After all, when you want to travel between two large cities in your country, you don't use the shortest route in terms of route. No, you'd rather use the high way - you can go much faster on a highway. And because highways run past hub cities, you'll take a bit of a detour, but only in terms of distance - you still save time. The analogy is sound, but for airplanes it is costs that are saved, not time.

### 47. Don't fact check us - we depend on it.

Except that we just learned that the flight from Johannesburg to São Paulo is actually flown. Perhaps, when that point was first thought up by flat-earthers, it was true. Then it would mostly mean that there was a lack of feasibility of long-range flights from south africa to south america.

Perhaps you remember that the sum of two sides of a triangle is greater than the third side (Euclid).

If you look on the 'flat earth map' shown here - euclidean geometry is valid on a flat plane - then you see a triangle between the three. You immediately see my point; the distance between Johannesburg-London plus the distance between London - São Paulo must be greater than that between Johannesburg and São Paulo. It doesn't make sense on a flat-earth map, either!There is but one good response, and I put it into a picture. If you're wondering, that's a fictional character from Stargate: Atlantis, a very intelligent, sceptical scientist. I liked the picture and I'm bad at memes, so this'll have to do.

### 48. Because you hadn't seen the point by now.

Another one, this time Santiago, Chile, to Johannesburg. I'll skip the great circle path, if you don't mind.

The JNB-SCL flight does indeed require a stop. In São Paulo. Because the flight is there now, we can immediately posit that there was probably either a lack of feasibility or the distance was too far. In fact, that issue seems likely because the distance is $9172$ km, according to google.

### 49. Physics is a lie.

The argument here is that because the distance from the sun to the earth is batshit crazy, the little distance it requires to the poles isn't enough to cause permanent winter.

So, what are reasons in physics that would explain this? Let's first list my brainstorm, and then we will look onward:

- The Poynting vector, the directional intensity of the energy flux by electromagnetic radiation, is at an angle to the surface.At the poles, the angle between the Poynting vector is at near $90^o$ to the surface normal, so the angle is near the minimum, meaning near-zero power.
- The flattening of the earth contributes to that by putting the angle even closer to the minimum.
- Refraction due to the atmosphere, like a lens.
- Most of the surface is water (snow), which reflects most heat away.
- Atmospheric conditions (e.g. faster convection loss of heat).

From what I can find, it is the first reason which is the main cause. We can argue why this is, too.

Why do you think it is colder during the winter? The distance to the sun isn't the reason. No, the reason is shorter days but mainly the angle between the incoming light and the surface normal.

### 50. Why is there a difference in the climate of the poles?

An interesting question in its own right. Mind you, there is no reason given why the flat earth model would by consistent, but that's common in this article.

Not surprisingly, there are a number of good reasons. The main one is that land loses heat faster than water does. As a result, the Arctic which is mostly the Arctic Ocean, loses heat the least fast. Antarctica, mostly land, loses it faster.

### 51. Let's repeat the previous assertion, because reasons.

Yeh, it's just the same point again. Why do the climates of the poles differ?

### 52. Why do places like Iceland, Canada or England have a nice climate?

Also, why don't those at the same latitude south don't?

The answer is simple; the Gulf Stream:

Simplified image of the hot/cold ocean streams. [bbc] |

I'm not so sure their assertion is true [Wikipedia/Sunrise equation]. My initial comparison between the longest day in the Northern Hemisphere versus the Souther Hemisphere shows that they're quite well matched.The Gulf Stream influences the climate of the east coast of North America from Florida to Newfoundland, and the west coast of Europe. Although there has been recent debate, there is consensus that the climate of Western Europe and Northern Europe is warmer than it would otherwise be due to the North Atlantic drift, one of the branches from the tail of the Gulf Stream.

The factors that can break the symmetry are probably the axial tilt and refraction of sunlight.

We'll consider the phenomenon a bit more in #54.

### 54. Similar, but with Dawn/Dusk

The lengths of twilight are, indeed, different in the two hemispheres. However, this can be explained by axial tilt and the fact that the earth's orbit is an ellipse [Brian B's Climate Blog]. Due to what is essentially conservation of angular momentum, it means that as the earth is further out it spins slower.

A lot more can be said about this, and there have been books written solely to calculating this sort of thing. I won't diverge much more here.

So, we see that once more they assert their alternative model doesn't have these problems without showing this assertion to be true.

What's more, they rely on an extremely oversimplified picture in the reader's mind. And then capitalising on this simplified picture, so as to make you doubt it.

If the mainstream picture doesn't fit, then

A lot more can be said about this, and there have been books written solely to calculating this sort of thing. I won't diverge much more here.

So, we see that once more they assert their alternative model doesn't have these problems without showing this assertion to be true.

What's more, they rely on an extremely oversimplified picture in the reader's mind. And then capitalising on this simplified picture, so as to make you doubt it.

If the mainstream picture doesn't fit, then

*realise that it is a simplified*one you were taught. Most of those things were taught to you during high school, and you lacked the necessary tools and knowledge to absorb the complete thing. What's more, it can't be expected that everyone knows everything.### 55. A rather silly picture.

Oh, this is so very, very wrong. So here, they present their model for solar motion, and claim it is merely a circle over part of the earth map.

They apparently fail to see that the sun - which is a ball of extremely hot gas with a hydrogen fusion reaction going on in its core, due to gravitational pressure - the sun doesn't shine in one direction like a flashlight. It shines in all directions.

The entire disc would be illuminated, the

*entire*time. It doesn't even explain day and night, let alone all the other things they claim it explains. Furthermore, if this were true, it would be easily measurable by angles - and it hasn't been measured.Yes, their model completely ignores all evidence to the contrary. That's sort of the point of my article.

### 56. Axial tilt doesn't exist.

The polar regions are interesting in two ways. First, we mentioned flattening. Second, there's axial tilt, which is explained quite clearly in the image below.

Image displaying axial tilt and the Arctic and Antarctic circles. |

Midnight sun can be explained by both of these. First, the phenomenon is possible because the axis is tilted with regards to the plane we move in. Second, the area is made larger than considered for a spherical earth because the poles are flattened.

Finally, one must remember that the sun is rather big - the approximation is a disc, not a point.

Taking all of these together, the phenomenon is completely explained [Wikipedia/midnight sun]. Why they ignore axial tilt is beyond me.

### 57. Denial of Antarctic midnight sun.

A bit of google shows that this is a big point for flat earthers. It must be hard when things can be debunked so easily.

First, they claim that they do not allow `

There's a fun way to show that they're wrong, both in denial of the midnight sun and that nobody can go there.

*independent explorers'*to go down there to 'verify or refute'. Perhaps safety concerns are in play?There's a fun way to show that they're wrong, both in denial of the midnight sun and that nobody can go there.

You see, there are commercial ventures that advertise with it:

Crossing the Antarctic Circle is an impressive achievement, as most expeditions to the Peninsula do not reach 66° 33’ S. You and your shipmates will celebrate in style with a well-earned glass of champagne! With a toast to the first explorers who ventured this far south, you can take pride in knowing you’ve made it to a part of the world still visited by very few people. This is raw Antarctica, a special area that you’ll experience during the midnight sun if you’re on our first voyage of the season.

From `Crossing the Antarctic Circle: Land of the Midnight Sun, days 9 & 10'.

### 58. The Belgians found this, therefore assertions.

A belgian expedition found that the sun sets on May 17th and rises only the 21th of July. This was at 71 degrees south latitude. They assert that this is '

*completely at odds'*with the sensible theory, without explaining why. You know, like in the rest of their entire article.
Let's first establish something. They said it was from

*71 degrees south latitude onwards*, but this is not true. The actual expedition got stuck/trapped in the Bellinghausen Sea, and had to remain there for the entire winter. As a result, they were stuck at that latitude, having frozen the vessel in ice for the winter. As this is within the Antarctic circle, it is not at all at odds with the sensible theory. Indeed, the Antarctic circle is defined as the latitude closest to the equator for which the sun can stay up for longer than 24 hours (or, indeed, down).Their assertions depend, as usual, on not showing that anything is actually true. In particular, it is noteworthy that they don't define the Horizon on their flat earth model.

### 59. Let's ignore Axial Tilt some more.

They now assert that the seasonal variations in the length of day/night isn't accounted for in the mainstream theory. This is rather silly, because one of the reasons it is so powerful is precisely because it does account for it. And teaches it to primary school children, I might add.

I made you an image to explain, using a globe I grabbed from google. It's not much different from the image we saw in #56, but shows more details we want to focus on. Mind you, it is a little bit skewed because the globe shows the pole. Still, it suffices.

A picture I made to explain the seasons and day/night length. |

So, let us first look at the seasons. As you can see I drew in the axis of rotation and the equator. Quite clearly, when you compare left/right, the position of the word 'the sun' is different with regards to the equator. In the left image, it is

*below*the equator, and in the right image it is*above*the equator. That explains why there are seasons, for instance because of the Poynting theorem we discussed at one point, that related the inclination angle of light to the (max) intensity absorbed. Wikipedia contains a rather nice animation which I recommend.
As for the length of day\night, this too can be readily seen in the picture. Draw the line that goes straight through

*the sun*. This line intersects the globe. You can see that the darkened areas vary with latitude; and that explains the difference in day and night lengths. Yes, it's actually that simple.### 60. Planks and tripods

So, here they actually describe an experiment. I'll describe it:

- Set up two tripods.
- Lay a (straight) plank over the two tripods.
- Look over the plank to the horizon, which aligns with the horizon.
- Observe that the horizon is flat if you move around the setup.

They also mention a level, I assume dumpy level, in the list of requirements but do not describe what they do with it.

They claim that if the Earth

*`were a globe'*then there would be an easily visible curvature of 66.6 feet (if the horizon over the plank measures 20 miles). I'm not sure what they mean by curvature here; the curvature I'm thinking of has dimensions of one over length.
Either way, let's think about this. The distance is $32.1869$ km, while the 'curvature', which I will assume is meant as the drop in horizon height moving from the centre to the right of the 'plank', is a mere $20.29968$ metres. The ratio between these is $6.30\cdot10^{-4}$. This means that the drop over the horizon is $0.063$ percent.

The length of the plank in their example is 6 to 12 feet, so at maximum the plank is $3.6576$ m. The horizon drop that you would then, in their example, be able to discern

The length of the plank in their example is 6 to 12 feet, so at maximum the plank is $3.6576$ m. The horizon drop that you would then, in their example, be able to discern

*with your eyes*is 2.3mm, which is certainly measurable. However, that's the horizon drop. So, if you align the edges of the plank with the horizon, the centre would rise 2.3mm above it, given their numbers.
It is at this point that I looked at the numbers. The described experiment immediately made me think of something, and it turns out I was right. It is the bedford level experiment, formulated differently. The ten miles to 66 feet was a give-away. The bedford level experiments, although I'd call them bets, are a number of experiments that were essentially wagers with flat earthers. The outcome is quite clear, although keeping track of the water's refraction is quite important. It is this latter detail that the flat earthers often try to use for their advantage. While wikipedia has a rather neat article on it, it lacks a clear conclusion. The introduction is, by far, the most clear:

But how far can this sort of thing go? Well, let's go back to the late 19th century:

As for the setup discussed in the article at hand, it probably is right. You know why? Because you simply can't resolve it. Let us, for a moment, assume that you are at low altitudes. What is the horizon?

So, when can you see the curvature? The answer to this question is not easily resolved. Then, the radius of this circle will be $3.57 \sqrt{h}$ [Wikipedia / Horizon]. From Applied optics, 47 (34), we can find a formula for the sagitta $S$, as it is apparently called. This is the 'drop' observed in the horizon. However, the formula requires the 'horizon' distance from some center point, which it calls $x$. How much is this? Their Fig. 6 shows the setup, and we just need to have the 'Field Of Vision' $\varphi$, which I've defined as the angle between the centre and the rightmost point in the field of vision. Using the definition of the sine and the distance to the horizon, we find $x = d \sin{\varphi}$. Let's put that in.

:

$$\begin{align}S\left(h,\varphi\right)&=R - (R^2-3.57^2 h\cdot\sin{\varphi})^\frac{1}{2}\end{align}$$.

I don't think I've had a surface plot yet, so here is a surface plot. Yay! As you can see, at the largest height I've plotted, ten kilometers, the apparent drop is a mere 12 mm when your field of vision is $\frac{\pi}{2}$. The field of view of humans is typically around $\frac{3}{4}\pi$, so $\phi\approx\frac{1}{2} \frac{3}{4}\pi\approx 1.2$ is where you want to look for a realistic estimate. That's around 9 mm if you're standing on top of an airplane. Looking out an airplane window, you won't be. At best, my camera would be able to get $\varphi = \frac{\pi}{6}$ (about $5$ km), I think. However, the sagitta has to be compared against the distance to the horizon to determine the angular sagitta, which tells us something about whether or not it is detectable. The Applied Optics paper, of course, did this, and determined that around $\varphi=\frac{\pi}{3}$, altitude $h=35\cdot 10^3$ feet, the angular sagitta is merely $0.51^o$. The angular resolution of the eye is listed throughout the paper as $1^o$. Their conclusion is thus that while it can be

However, I am not entirely sure about something specific in this this paper. There seem to be some things wrong; in particular, conversion from the Sagitta in [km] to angular dimensions $\sigma$ is a bit easy. Merely dividing it by the distance to the horizon yields, I think, the $\tan{\sigma}$. So you need to apply the arctangent function to that ratio. Mind you, the first order approximation to that function is the ratio itself. Using that as an assumption and multiplying by $\frac{180}{\pi}$ to convert to degrees and by $60$ to convert to arcminutes, you find these results:

What I find particularly illuminating is that my own version allows us to tell when it is visible. The Field of vision required is $\varphi=0.46$ [rad] and the height is a mere $1200$ km. Which means that yes, it would be visible at moderate heights. Mind you, $\varphi\approx 23^o$, so the total vision is an angle of $46^o$. The camera used can get to about $40^o$, based on my estimate and compared to the values I find on google. To the right we can see the sagitta in arcminutes for this estimate of my camera. As you can see, for most cameras the angle won't be visible until you're a bit further up. Such a camera would, I estimate, have to go up to about $14$ [km], which is not done on regular flights. Many amateur projects involving balloons and such do go up that high, which is why they can photograph it using a GoPro or similar. In any case, a person that moves his head around can get a far wider field of vision, and would thus be able to see the curvature.

It's actually a story used quite often by flat earthers, although they loathe naming it now because the rebuke is so readily available. Especially the Rational Wiki resource is becoming quite widespread....later attempts to reproduce the observations firmly support that the Earth is a sphere.

But how far can this sort of thing go? Well, let's go back to the late 19th century:

This is the proud heritage of the Flat earther. If you're interested in reading about this, I suggest you read the Scientific American (Blog) on it.Mrs. Wallace,—Madam, if your infernal thief of a husband is brought home some day on a hurdle, with every bone in his head smashed to pulp, you will know the reason. Do you tell him from me he is a lying infernal thief, and as sure as his name is Wallace he never dies in his bed.You must be a miserable wretch to be obliged to live with a convicted felon. Do not think or let him think I have done with him.John Hampden.

As for the setup discussed in the article at hand, it probably is right. You know why? Because you simply can't resolve it. Let us, for a moment, assume that you are at low altitudes. What is the horizon?

So, when can you see the curvature? The answer to this question is not easily resolved. Then, the radius of this circle will be $3.57 \sqrt{h}$ [Wikipedia / Horizon]. From Applied optics, 47 (34), we can find a formula for the sagitta $S$, as it is apparently called. This is the 'drop' observed in the horizon. However, the formula requires the 'horizon' distance from some center point, which it calls $x$. How much is this? Their Fig. 6 shows the setup, and we just need to have the 'Field Of Vision' $\varphi$, which I've defined as the angle between the centre and the rightmost point in the field of vision. Using the definition of the sine and the distance to the horizon, we find $x = d \sin{\varphi}$. Let's put that in.

:

$$\begin{align}S\left(h,\varphi\right)&=R - (R^2-3.57^2 h\cdot\sin{\varphi})^\frac{1}{2}\end{align}$$.

Following the Applied Optics paper, this is in angular dimensions. |

*measured*from a (proper) photograph, it can't be*seen with the naked eye*. Indeed, the proper photograph they show, Fig. 5, does not allow me to resolve the horizon with my eyes, until I cheat and zoom in.However, I am not entirely sure about something specific in this this paper. There seem to be some things wrong; in particular, conversion from the Sagitta in [km] to angular dimensions $\sigma$ is a bit easy. Merely dividing it by the distance to the horizon yields, I think, the $\tan{\sigma}$. So you need to apply the arctangent function to that ratio. Mind you, the first order approximation to that function is the ratio itself. Using that as an assumption and multiplying by $\frac{180}{\pi}$ to convert to degrees and by $60$ to convert to arcminutes, you find these results:

Sagitta [arcminutes] versus height, field of vision and the 3d plot. |

Sagitta [arcminutes] versus height for my (estimated) camera. |

Photograph taken at 30000ft, flight from Amsterdam to Geneva |

As you can see, it is barely visible in the best of cases. At heights below the $4000$ m ($13000$ ft), I wouldn't bet on seeing it. The article in Applied Optics also describes some other obscuring effects. So, what'd it look like? Let me pull from my own images (for once). I took it out of an airplane flying from Amsterdam to Geneva, at a height of 9.2km ($30000$ ft). As you can see, it is barely visible. As the Applied Optics image outlines, there is a slight skewing because my camera isn't well aligned to the horizon. I could see it, just barely, but I am also wearer of glasses. So I can't entirely trust that observation. Indeed, comparison to Fig 5. of the Applied Optics paper shows us that my camera is affecting the results, and likely my glasses are as well.

As a final remark, someone from my study had something cool quite recently. He's on the TU Delft DARE project, and they launched the Stratos II+ rocket. It got to quite a high altitude. But what's relevant here is that it is a rocket with mounted cameras on it - so you can see around it when it is ascending. For that purpose, I link to the youtube video, slowed, here. Youtube slow is a website I've used before; it allows you to watch it slowly. The rocket is spinning quite rapidly, which should help it stabilise. Note that the left lens does seem to be a non-fisheye one, but I can't say for sure.

In conclusion, apparent flatness isn't equal to actual flatness. There are better methods available to show it. We've known since

*at least*the ancient greeks that the Earth isn't flat.

### 61. Buildings should tilt backwards.

They probably do, at that. But by how much? They also claim that the horizon should be noticeably curved at sea-level, but we've already discussed that in #60. So let's calculate the backwards tilt.

This one can be done pretty simply. The distance along a sphere is simply the (radian) angle of the curve, times the radius. We've already seen the distance in km as a function of height in m. So, we can calculate the angular difference rather simply:

$$\begin{align}d\phi &= \frac{3.57 \sqrt{h}}{R}\end{align}$$

If you're wondering, that ends up in $5.6035 \cdot 10^{-4} \sqrt{h}$. So, standing on the top of the Big Ben (96 m), you would be able to resolve.. $0.0055$ radians or $0.3146^o$. We've already discussed the angular resolution of the eye, which is $1^o$. You would be able to resolve such a thing when you're at about a kilometre high.

The tallest building in the world is the Burj Khalifa in Dubai, but it is only $828$ m high. Can you see it? I think I can, but I'm not sure.

### 62. Bedford Level Experiments

We've already discussed this, but it is of course clear that the result is in favour of a non-flat earth.

I'll point to the scientific american article again, and include this quote:

This new setup showed the same thing as the first: the earth was indubitably curved. No reasonable person could doubt it. Alas, Wallace was not dealing with reasonable persons. They responded in true creationist fashion: by completely refusing to deal with reality.

### 63. Bedford level, different one.

Yes, another one. Given that there is no source, and it is another case of Rowbotham, who as far as I can tell isn't a Dr., yelling that he did an experiment and found what he was looking for.

I find it far more telling that, whenever such an experiment of Rowbotham was held up against sceptics, it was found that the earth curved.

I find it far more telling that, whenever such an experiment of Rowbotham was held up against sceptics, it was found that the earth curved.

### 64. More Rowbotham.

This is exactly the same as #60, the plank experiment. This time, instead of a plank it is a piece of string. As we've already discussed at length, you can't observe the horizon being curved below a rather high height.

### 65. More, more Rowbotham.

Another experiment, similar to the first Bedford level experiment. Basically, he watched a steamer, which is probably a large steamboat, `sail' towards the horizon. Because it is hilarious, we will use a flat-earth video. It should be at #t=802. Look closely; he will start comparing the two times. He apparently claims that this is atmospheric refraction, not the horizon. However, look at the bottom of the ship! It is clearly disappearing below the horizon. So, an image that shows a ship disappearing behind the horizon is used as evidence that it doesn't.

*for*a round earth, with someone called Red's Rhetoric filming it.As for angles, we already discussed these. You probably won't be able to see this until you're at a height of about a kilometre.

### 66. More of not-a-Dr. Rowbotham

We won't discuss more of his level experiments; we know that those examined by scientific scrutiny showed that the earth is curved.Here, they claim N.D. Rowbotham "

*caused quite a stir in the scientific community"*and "*the shape of the earth became a hot topic of debate around the turn of the 19th century"*. Of course, this didn't happen. Wikipedia informs us that a pop-science magazine spoke had it as a regular feature. The scientific community at large wasn't impressed or involved.No, what it mostly did was this:

... Hampden, who had embarked on an extraordinary 15-year campaign of abuse and libel that landed him in both jail and court several times. [Scientific American Blog]

### 67. Light bends.

Let's reverse our formula for horizon distance. The horizon you see is the lowest point. So, how far away can we see the Great Orme's Head? Great orme has an elevation of $207$ m. The hill has an elevation of $30.48$ m. As the object far away is higher up, you can see it better. Intuitively, I surmise that we sum the heights. Thus, we have $h=237.48m$ which leads to a distance of $55$ km. Clearly, we aren't able to see it directly - something is interfering.

Still, Wikipedia notes that it is visible, although they clearly say that it is the summit, not the head:

This might be the right place to talk about what is interfering. Also, we'll edit our distance formula slightly, taking into account refraction [Good description here].Due to its location, the Isle of Man and the Lake District can be seen from the Orme's summit on clear days.

Mirages are optical phenomenon due to the different densities of air causing different bendings of light [Wikipedia/Mirage]. A well known mirage effect is that the sun appears flattened when setting. They are the better known one, but this sort of thing happens all the time. Looming is fun too [Wikipedia/Looming]. I first saw that in the plane to London, when I saw a lot of boats forming a sky armada. Rather than believing my eyes and believing the Empire was about to strike, I thought about refraction for a bit and continued to read my book. The green flash is good evidence, too.

I won't redo the calculation, but it has been done. The estimate for the horizon distance becomes $3.86\sqrt{h}$, due to light bending. There's a lot of assumptions going into that, largely because refraction depends on atmospheric results - basically, the weather.

*On a clear day*doesn't mean that the weather is nice, but is actually a reference to the optimal conditions for far-seeing. We can see that described nicely [Wikipedia]:Well, that makes sense.When conditions are unusual, this approximation fails. Refraction is strongly affected by temperature gradients, which can vary considerably from day to day, especially over water. In extreme cases, usually in springtime, when warm air overlies cold water, refraction can allow light to follow the Earth's surface for hundreds of kilometres. Opposite conditions occur, for example, in deserts, where the surface is very hot, so hot, low-density air is below cooler air.

I don't know the particular atmospheric conditions that are present on '

*a clear day' ,*i.e. a day on which you can see it, but I propose that refraction is the answer.Denial of refraction is a hard one. Above, I've already pointed to a few points of evidence. Other things that depend on it are telescopes, glasses, microscopes and cameras. Probably a lot more, too. The underlying theory of light is so very, very settled by now that I can't imagine someone denying it. Of course, even if I can't imagine it, I know someone will, anyway. The underlying theory is the wave theory of light, from which Snell's law can be derived. And below that lie Maxwell's equations. Which are consistent with quantum mechanics, and on which a lot of work is based.

### 68. More refraction

This time, they don't even mention atmospheric conditions but simply put down an image. Well, that's cute. I look at that horizon and I see various discolourings. Since the refractive index depends on wavelength, that's a clear indication of refraction in the distance. To me, that's sufficient.

Note that the drop they discuss is only $100$ m. We'll mention this point again, but it is called a

*skyline*for a reason.### 69. You asked for refraction? Or is tall buildings being tall sufficient?

So, here we are again. The distance is 60 miles, the elevation is from Bear Mountain ( $1283$ feet). Among the clearly visible buildings is the Empire state building ($1250$ feet), I think?

It doesn't really matter. The height of Bear Mountain allows for a horizon at $76$ km. However, what we forgot to include is that New York City is at 500 feet elevation itself.

Let's use my intuitive guess from earlier and just add those. That gives us $89.9851$ km, which still isn't enough.

So yes, part of the

*tall buildings*will be hidden. That's why it is called a*skyline*. Because it's the upper part; the bottom of the buildings is cut off. The Empire state building is $381$ metres high - the $52$ m drop they discuss still allows you to see it.A quick calculation using the height from Bear mountain, the 500 feet elevation of new york, and inversing the horizon distance formula, tells me that any building higher than $124.2365$ m is visible. That leaves

*more than a hundred buildings in new york*that are visible [Tall buildings in New York].### 70. On a clear day, refraction.

They took their time to mention that we're talking about

*a clear day*. Note also the discolouring on the image. I'm not going to spend much more time on pointing out refraction.
You know what? We will describe how to falsify the refraction explanation.

Requirements:

- Three monochromatic high-power lasers; preferably, one red, one green and one blue, because primary colours. Ultraviolet lasers are also an option.
- A frequency generator of sufficient power.
- Three synchronised detectors.
- Other stuff that's not important enough to list.

If you're wondering about the frequency generator and synchronisation, it is because of a trick that allows you to detect dim objects.

So, what is the point? Well, we know that the refractive index depends on the wavelength. A widespread of lasers allows for monochromatic light of different wavelengths, so that one can directly falsify or confirm that prediction.

Ideally, the setup is such that the lasers are aligned in parallel from New York City to Bear Mountain. With a lot of effort, one can place the detectors such that they detect the lasers. If you're afraid of a fake signal, use a music song to regulate the laser. Yes, that can be done, and while it complicates the data analysis it is also extremely powerful as a way of detection.

What will you find? Well, if the refractive theory is true, then the differently coloured lasers will have detectors at different locations. It's that simple.

So far, this has been confirmed. The essence of the experiment isn't that hard - it basically uses lasers to determine the refraction, which can depend on various atmospheric conditions. And would one find this?

Perhaps:

So far, this has been confirmed. The essence of the experiment isn't that hard - it basically uses lasers to determine the refraction, which can depend on various atmospheric conditions. And would one find this?

Perhaps:

Atmospheric effects deleteriously impact free space laser communications. Beam wander, distortion and beam bending can affect pointing and tracking in particular. Mirages are an example of these effects. In June 2006, a campaign was conducted across the Chesapeake Bay by the Naval Research Laboratory to quantify effects of mirages at the marine layer. We imaged a series of lights positioned strategically on a tower across the bay, at Tilghman Island, approximately ten miles away from NRL's Chesapeake Bay Detachment (NRL-CBD). Recorded images were subject to displacement and distortion as functions of temperature, humidity, dew point, and other meteorological parameters. Results from the experiment will be presented and phenomenology discussed. [doi:10.1117/12.800782]

### 71. Something about mirages not behaving like they do.

So, here they claim that mirages are 'hazy' and 'illusory'. Also, apparently they think mirages require the object to be upside down?

As we've discussed, mirages happen due to refraction bending the path of light rays. They do not require inversion. The haziness is mostly the different wavelengths, and is dependent on the background. The picture they show fits 'haziness' perfectly, or more properly, discolouration.

As we've discussed, mirages happen due to refraction bending the path of light rays. They do not require inversion. The haziness is mostly the different wavelengths, and is dependent on the background. The picture they show fits 'haziness' perfectly, or more properly, discolouration.

### 72. Mirages don't happen.

So, here they continue elaborating on how they ignore mirages. Given the events they quote, I can't really verify if the anecdote is true.

It is certainly true that the object is normally hidden behind the Horizon. So we assume it is a mirage.

It is certainly true that the object is normally hidden behind the Horizon. So we assume it is a mirage.

### 73. More mirages

Another anecdote, and I'm not going to bother finding a source for it. These anecdotes usually lead to a few hundred copies of the 200 ``proofs''. Either way, the distance is such that the top of the island would be just barely visible.

So, mirage. It happens - deal with it. Again, note the

*on a clear day*.### 74. Another mirage over a large body of water.

What a surprise.

### 75. Genoa has mirages often.

And this one gives us a Youtube movie too.

### 76. Yes, Genoa has mirages.

What did we just say? Elba is in the video.

### 77. We know...Genoa and mirages.

Yep, it's in the video.

### 78. Yes, Genoa.. Wait, Anchorage?

Okay. First, there's a very clear superior mirage in that image - the foot of the mountain is not visible, but there is a different colour there (etcetera).

Also, it happens quite often there - the latitude and the geography combine to regularly cause mirages. E.g. the Eleutian range is often visible.

Also, it happens quite often there - the latitude and the geography combine to regularly cause mirages. E.g. the Eleutian range is often visible.

As for the angle, we've already discussed it. It's not much, although this height should be resolvable. Now, I'm not familiar with these mountains, and mountains are known for having so many ridges and angles that I really can't tell if it's straight up or tilted.

### 79. Anchorage, again

Inventing new arguments is hard. Different mountain, same story.

### 80. Mirages can be seen by sailors, too.

Mirages can be seen by anything that can see. Don't be surprised if sailors see it.

### 81. Mirages and lighthouses.

You know, we've discussed this.

Let's just say that the photographic evidence tells us that these are mirages.

### 82. Example of a lighthouse.

### 83. Another lighthouse.

### 84. Hey look, a lighthouse.

### 85. Had we done a lighthouse yet?

### 86. More lighthouses.

### 87. I'd really like to discuss a lighthouse now.

### 88. Nobody expects.. a lighthouse?

### 89. We are the lighthouses that say...

### 90. Nobody expects.. the statue of liberty?

This is not a lighthouse. And

*on a clear day*, i.e. a day for mirages, it is visible from far away.
I'm going to make another prediction; using filters to show something is a mirage. We already know that the refraction index is dependent on the wavelength of light, so you can probably show something is a mirage by taking monochromatic pictures.

I'm quite optimistic this can be done. I can't find anyone that

I'm quite optimistic this can be done. I can't find anyone that

*did, though.*

Otherwise, a polarisation filter should be able to remove the mirage, because those rays are usually bent.

### 91. Finally, we can talk about lighthouses.

### 92. Sir, we've discovered a Mirage Denier.

Yeah, I'm getting pretty tired of this. It'd be pretty amazing if there was a flat earther that didn't go full confirmation bias on us. I've seen mirages, scores of them - as has everyone else, but most people don't notice. On the other hand, I'm the sort of person that's on the watch for it. So I saw a few floating ships, looming mountains in geneva (in retrospect - I've never seen mountains before. I'm from the Netherlands.) I've been noticing it on the roads, rather often.

### 93. Lighthouses!

The holyhead breakwater lighthouse has a regular range of 23 km, and a height of 19m.

The other one is similar; 20m high.

The other one is similar; 20m high.

The rest of the difference can probably be explained by a mirage.

### 94. Let's get back to something refuted earlier.

The horizon seems flat!

Yes, that's why we discussed the Applied Optics paper.

Yes, that's why we discussed the Applied Optics paper.

### 95. Let's do that again.

### 96. Lighthouse Lasers

Apparently, lighthouses shine with lasers, or something. Or not even a lighthouse; but just 'the light of'.

Well, light is usually a spherical wave.

Well, light is usually a spherical wave.

(I can't actually be bothered to check this anecdote.)

### 97. Science doesn't evidence, or something.

Apparently, the observational evidence for the big bang doesn't exist. Neither does that for the milky way. Or for its motion. Of course, neither does that for the sun or basically anything else, except the flat earth and unicorns.

### 98. Absolute numbers are impressive.

Did you know that? The most accurate distance measured is 430 +- 30 light years.The error is still large, but that's probably because of limits on the measurement. Such as measuring from inside the atmosphere.

### 99. Had we ignored axial tilt in the last 10 reasons? Time to ignore it!

This is basically a statement that polaris can't be seen below the equator. But it can be, because of axial tilt. Not much of a surprise there.

### 100. Milestone! Also, what do they mean?

I'm not fully sure what they mean here. I think they're sort of imagining a ceiling of stars with a sphere below it or something?

You can't see all stars simultaneously, because they are

*all around us*. And one side of the planet is*day*while the other is*night*.### 101. Let's bash Brazil's flag.

So here they claim that there is `

It's on the flag of Brazil, which is a hint to how legitimate the speculation is.

As for

As for '

*legitimate speculation'*whether sigma octantis exists. No surprise there; the star is very dim. You'll need some equipment to see it, which you can of course do.It's on the flag of Brazil, which is a hint to how legitimate the speculation is.

As for

*'cannot be seen using publicly available telescopes',*it is of the same apparent magnitude as Uranus, which can be seen by both the naked eye and telescope, even in a Urban Area. It was easier with the telescope, though. The naked eye can discern up to +6.5 magnitude, while octantis is at +5.42. It is visible.As for '

*not motionless',*Wikipedia tells us:To an observer in the southern hemisphere, Sigma Octantis appears almost motionless and all the other stars in the Southern sky appear to rotate around it.

### 102. Maths are hard

So, here they tell us that the angle of Polaris in the sky is due to 'the law of perspective', which dictates that the angle and height diminishes the further one recedes (on a plane).

Cute, but you can actually calculate the rate at which this would happen for both of them. Guess which one it is.

### 103. Please, ignore Earth's tilt. A completely novel argument.

Via a star map. |

The argument here is that Ursa Major is 'very close' to polaris, but visible down to 30 degrees south latitude. (Polaris is visible to 23-ish.)

Close, of course, is a relative term. It's actually quite a bit away, as we can see in the image to the right. The angle difference is roughly 13 degrees, which mean's I'd expect ursa major to be visible down to roughly 36 degrees south latitude.

It's important to note that the earth is at a tilt, especially for astronomical purposes. The same, of course, can be said for celestial navigation purposes.

Watching the clear night sky can be quite fun. I still can't tell one constellation from the next, but I got pretty good at spotting Mars last year.

### 104. Spheres are a hard shape to see.

Really? This constellation is at declination +25. `Logically' I'd say it is visible from $25+90 = 115$, or the entire hemisphere, down to $25-90=65$, or most of the inhabited southern hemisphere. (With a really rough estimate).

Do you want a picture, or what?

A really bad sketch of the situation. Note the green and blue sight fields. |

So no, that actually makes perfect sense. Now try explaining that by

*your flat-earth model, in particular the specific*### 105. Let's assert our model has all the solutions, except alternatives for all of physics!

So a list of similar objects to #103 and #104 follows, after which it is claimed that there model - which is apparently communicated in pictures of at most the quality of my rough sketch, above - that their model has no such problem!

Of course, there isn't even a problem.

Of course, there isn't even a problem.

### 106. Apparently, there is no south pole.

So, here they make the mistake of going for a magnetism argument. The magnetic north pole isn't the true north pole, so why do they expect that from the south pole?

Additionally, as far as I know nobody has wanted to check for the magnetic south pole in a frozen wasteland. Surprise.

The geomagnetic poles do move around; this isn't an

*establishment*claim (oh no, big government! Mainstream! Noes!) but a scientific one. Yes, it a scientific theory, which means that it has the evidence going for it. I hope that amount of links makes my point? Good.### 107. Magnets, how do they work?

Let's start with the core model. I know from signal analysis that part of the evidence going is essentially earthquake transmission. There is quite a long story.

As for ring magnets, yes, you can make ring magnets that have a dipole moment that is a radial vector. You can also have one with the cylindrical z-vector. It's probably possible to have one with the polar vector, at that.

Now, does that mean their 'model', which hasn't been presented, ever, is a better one?

No, not really. For instance, because the field lines would be measurably different from that of the earth model.

No, not really. For instance, because the field lines would be measurably different from that of the earth model.

### 108. Timescales are a myth

You can make magnetic field lines visible by dropping iron fillings around the magnet. These can be thought of as really tiny compass needles, that point along the field lines. |

The movement of the magnetic poles has a sufficiently larger timescale than that of a sailing trip. Or several of them. Or a thousand years of them.

Furthermore, a compass doesn't point towards the pole. It points along a field line. A field line isn't an actual line, but it is a good way to show them. Essentially a magnetic field is a continuous variable, but if you want to draw them you use field lines. Iron fillings, such as depicted to the right, turn into tiny bar magnets themselves. This means they have their own field, so the filings close to each other line up to the effective field, which is simply the sum of that of the other filings plus the big bar magnet's field. The reason they do line up is simply minimisation of energy - the driving principle of most of classical mechanics, but also of quantum mechanics, thermodynamics, and basically most of physics.

### 109. Here's something that's actually sort of true.

The concentric circles thing and the 'handedness' of the course is true for both a spherical and a flat earth. Except, of course, because of tilt.

They still have to explain, well, anything that's in their model. Perhaps you should describe the actual model and then post it for peer-review?

### 110. Apparently, distance is a myth, too.

They do realise how much longer it would take at different latitudes, in their non-existent model?

No? Ah. Also, it's not often quoted as proof, because it's just a demonstration. You do realise we've known that the earth is roughly spheroid since the Ancient Greeks, possibly before that?

### 111. Apparently, polar routes don't happen.

Of course, as we've already discussed earlier, they do happen.

And no, it is not standing proof of earth not being a ball. It's standing proof of great circle routes not often going over the poles, and of people not liking the cold much.

### 112. The earth doesn't rotate or something?

So here, they seem to say that because the earth spins around the sun, so day/night have to flip every 6 months. I'm a bit confused about what they're trying to say; they know the earth spins around its axis, right?

### 113. People fall off the earth in Australia, or something.

There is a direction of 'up' and 'down', yes. It is either parallel or antiparallel to gravity. Simple, really.

### 114. Because ignorance has to be worn proudly.

Pretty much the same, but now with some third and fourth century guy who didn't know up from down. Gasp, and all that.

### 115. Science doesn't work by evidence after all.

I'll quote this one:

First, since it predicts planetary orbits very well, and led to the discovery of a new planet when it didn't fit, it's quite well established even without direct experimentation.Not a single experiment in history, however, has shown an object massive enough to, by virtue of its mass alone, cause other smaller masses to be attracted to it as Newton claims “gravity” does with Earth, the Sun, Moon, Stars and Planets.

You do realise airplanes, rockets, satellites and various other things are built using Newton's law of gravity? Good.

So here's the Cavendish experiment:

You know, it's generally a good idea to keep up to date. Using arguments that haven't been valid since before the bedford level experiments - arguments that haven't been valid for more than 200 years doesn't bode well for your credibility.The Cavendish experiment, performed in 1797–98 by British scientist Henry Cavendish, was the first experiment to measure the force of gravity between masses in the laboratory and the first to yield accurate values for the gravitational constant

Note that the Cavendish experiment has been reproduced a great number of times.

### 116. Satellites are a lie.

Apparently, the

*magic theory of gravity*has never been proven. Except, of course, by experiment? By planes? By springs, slings, waves, projectile motion, satellites, rocket motion, the motion of planets, and so on and so forth?
The equation isn't hard, and I mean it. The acceleration required on a particle to be 'locked in orbit' for most objects is given to it by gravity. The equations are:

$$\begin{align}G\frac{m_\text{Earth} m}{r^2} &= \frac{m v^2}{r}\\G m_\text{Earth} &= v^2 r\end{align}$$ for an object of mass $m$, velocity $v$, distance to earth's centre of mass $r$. $G$ is the Newtonian gravitational constant.

$$\begin{align}G\frac{m_\text{Earth} m}{r^2} &= \frac{m v^2}{r}\\G m_\text{Earth} &= v^2 r\end{align}$$ for an object of mass $m$, velocity $v$, distance to earth's centre of mass $r$. $G$ is the Newtonian gravitational constant.

It's not magic. I think Dr. Thijssen shows it in the video we started it, but you can even check if the exponent is correct for the motions we find (it is). You can check if extra terms are needed (they aren't). The reason we hail Newton's name among the giants of physics, is that he formulated these laws, combined the existing pieces of mathematics into what we now know as calculus, and used this to explain Kepler's observations.

### 117. Passing mathematics isn't mandatory.

Apparently, because this guy sure did not. Gravity depends on distance and on masses. And, surprisingly, mass doesn't include only volume. To illustrate that point, there's the old question; what's more massive, a kg of feathers or a kg of lead? Well, both are a kg, so they are equally massive. Which one's got more volume? The feathers, of course.

The difference in distance between the two opposing sites of the earth is two times the earth's radius. The force of gravity is inversely proportional to the distance squared; there is thus a difference between the force acting on water on the far side and water on the near side of the moon. As a result, water flows.

The theory of tides is slightly more complicated than that, where this looks like a neat explanation. Read it if you wish.

The question you must ask yourself is this. Place a marble and a football so that their equators lie in a plane. What is the distance from the equator of the marble to the centre of the football? Is the distance from the top of the marble larger than that from the equator? The distance between the centres shall be called $L$. If the marble has size $x$, then the distance between equators is simply $L-x$. However, the distance from the top is a horizontal $L$ and a vertical $x$, so Pythagoras' theorem tells us that this distance is $\sqrt{L^2+x^2}$. As you might imagine, these distances differ.

The theory of tides is slightly more complicated than that, where this looks like a neat explanation. Read it if you wish.

The question you must ask yourself is this. Place a marble and a football so that their equators lie in a plane. What is the distance from the equator of the marble to the centre of the football? Is the distance from the top of the marble larger than that from the equator? The distance between the centres shall be called $L$. If the marble has size $x$, then the distance between equators is simply $L-x$. However, the distance from the top is a horizontal $L$ and a vertical $x$, so Pythagoras' theorem tells us that this distance is $\sqrt{L^2+x^2}$. As you might imagine, these distances differ.

Which mean the forces differ.

Sorry for the bad joke. Either way, that's how this reads. Let's just focus on this part:### 118. Let's claim something demonstrable false, and then make a point of something clear.

First off, the velocity and path of the moon are anything but constant/uniform.

And of course closed surfaces don't experience clear tides, the water can't flow. Still, the Great Lakes do have tides.

The LHC actually corrects for the tides.

### 119. NASA's got a bigger telescope than I do.

Basically that's their argument; if you take an amateur telescope, Mars and others are

*merely discs*.
However, hardcore amateur star gazers have big telescopes too. And.. they can see the shapes of the planets, too. Mars always grabs people's attention.

I'm not even going into 'I say earth is a plane, so it doesn't matter what planets are'.

### 120. Ancient greek is older than latin, thus latin.

Guys? Ancient greek predates Latin. Plus, plane comes from planum. The "I call it earth plane" argument isn't going to fly, not even if you google "Word etymology".

Ancient greek is the clear winner here, and it called wandering stars ἀστήρ πλανήτης. The first word, aster, means celestial body (or what they just called a star, back then). They did clearly see the differences, because they added planetes, from planeo, "I wander".

### 121. Is this a cellar? There's so much whine.

You can clearly see they are the same size and distance, you can see the Earth is flat, you can feel the Earth is stationary, but according to the gospel of modern astronomy, you are wrong and a simpleton worthy of endless ridicule if you dare to trust your own eyes and experience.

Two random people's vacation photo, taken from google images. |

*extremely fancy*. I googled for photo giza. Do you believe this couple is as large as the Pyramids of Giza?You don't? Why not? Because your

*experience*tells you that*your eyes are wrong*because far away things tend to look smaller.You are worthy of endless ridicule if you keep an idea when it is so very, very clearly refuted by common sense, daily experience, high school level physics (and beyond), and so on. You literally have the option of looking at a photograph of the Earth. But no, an elaborate conspiracy involving basically everyone that doesn't believe you has to be present because you want to

*trust your eyes and lack of experience*. Well, you can do that; just don't expect us to join you.### 122. Honestly, high school physics again?

Newton's first law. If there's no force, then it is a valid inertial frame in which the laws of physics hold.And not only that but then people would say, 'oh then how do you explain a fixed, calm atmosphere and the Sun's observable movement, how do you explain that?'

If that's so, show us your model. Show us the alternative hypotheses you have that are utterly perpendicular to everything we've learned over the last few centuries, and the millennia before that.everything can be explained by a motionless Earth without bringing in all these assumptions to cover up previous assumptions gone bad.

Show us these, instead of showing us silly pictures. Explain, predict, and calculate, instead of asserting and moving on to the next bit of your gish gallop.

### 123. People change their answers because of new evidence...

And somehow, that's a bad thing. Well, it would be if you believe in a model that's been outdated since before the first Greek asked for a towel (Eureka!).

Kepler didn't decide that, anyway, because it's not a circular orbit. He's the one that came up with orbits being more elliptical than spherical, so he wouldn't have said "It's this much". He would've given the two parameters to define the ellipse.

Again, I must point out that we went to the moon. This can be verified by various methods, you don't just have to believe it. Although the Soviets admitting it would be a pretty strong argument in and of itself.

Also, how does your flat earth thing, which so far hasn't explain anything, explain the super moons?

### 124. Reflection doesn't happen.

So here, we see a picture of what they claim is 'a hot spot'. I'd say it's more clearly a reflection spot. This would require a local area of the correct angle such that the light rays are reflected into the camera. They have not, at all, tried to show that this is a hot spot. It's a yellowish spot on the cloud layer, that much is visible.

We can also determine that they went out of their way to select a shot without the source. I don't know about you, but it seems to be similar to the solar reflection on the sea during sunrise/sundown, when it is near the horizon. The situation here is similar, but instead of the sea it's a cloud. It's still a sunrise, so this seems very likely indeed.

### 125. Clouds are, apparently, flat?

Another bad sketch, showing a sun that is close. |

Here, it is claimed that because the light rays visible through a cloud layer, always a very aesthetically pleasing effect, indicate the sun is close because of the convergence of light rays.

They do, by the way; at a spot slightly higher than shown in their tracing.

But where do they merge? Well, simply at the apparent location of the sun. Remove the clouds, and the spot of convergence is where you see the sun.

Does this show that the sun is close? No, it doesn't. The rays you see are at an angle, an angle pointing away from you. There's a lot of rays you don't see.

More bad sketches. You'll have to click this one. |

As you can see, a lot of rays won't even make it to the ground. This is because the sun, being so close, doesn't have rays parallel to the 'tunnels' through the clouds.

Now, what would it look like with a far away sun? Again, the sketch is to the right. You see that because of the distance from the sun, it has a lot of rays that are sufficiently parallel to the tunnels to go through. At ground level, the rays are quite a bit away from each other. The final question is: What would happen if you add perspective?

### 126. A false claim about what science claims about the seasons.

Because of the axial tilt and the orbit around the sun, the apparent 'height' of the sun changes throughout the year. For the norther hemisphere, when it is summer, the tilt is towards the sun, so that the apparent latitude is higher (i.e. the earth-sun plane is above the equator during day). In the winter, it is pointing away and the apparent latitude is lower (i.e. earth-sun plane below equator during day). It's not a very hard explanation, and we already saw it in #59.

So no, the mainstream model doesn't claim the distance to the sun is what determines the seasons. This is simply false, and only serves to shows that the 'mainstream model' is not understood by the writers of the flat earth arguments.

### 127. Reflection is difficult.

The straight line reflection between the sun and the observer is a simple result of possibilities. If you consider the water as 'level' with respect to the curving surface, then the only rays that can reflect are those that lie on a line towards the sun. If they're not exactly in the middle, then small waves down to microscopic levels reflect sunlight towards you. If they're too far to the sides, there's just no chance that light reflects towards you.

As for the 'straight line', have a simplified sketch again.

The curvature is very flat, compared to what most people would think. Still, it does affect the angle of incidence and it is the reason that you have this long stretch of reflection.

A flat earth with a sun that is close, and especially the sea as the surface, would act more like a mirror. While I won't tell you to grab a mirror and refract the sun in it, you can use a dim light and do the same. You don't get this line, but a reflection spot and that's it.

### 128. Timescales are a myth, too. So is precision.

Really inaccurate clocks are about as inaccurate now as when they were

The motions 'claimed' by modern astronomy are observations that have been explained. It's pretty simple; the data is out there, and one can verify it. However, as we've previously seen, this is just

*centuries (gasp!)*old. Yes, they aren't very precise, are they?The motions 'claimed' by modern astronomy are observations that have been explained. It's pretty simple; the data is out there, and one can verify it. However, as we've previously seen, this is just

*going too far*for flat-earthers; you must only do*your own research*if you don't actually have to do something.
Yes, spending a few hours doing

*your own research*sounds far better than spending about 5 years in university, so that you can spend 4 years as a PhD student, a few years as a PostDoc and then moving on to, finally, become a scientist.### 129. Some random person said something.

Why would you fics the telescope on a solid stone base?

One thing that comes to mind is that solid stone is solid. You can't accidentally move it, and it can't be moved (or not easily) by natural causes, such as (weak) earthquakes.

Most amateur telescopes are mounted on tripods, and we put the tripods on a hard surface. Not because of 'not moving a hair's breadth' - which is important with a very limited view angle - but because it needs a solid base. If you put in damp soil, for instance, then it will slowly start sinking into it. The effect is more pronounced when not all three of the legs are in equally damp soil. Note that a telescope usually has a counter weight, so that even a relatively small one like me has total mass of 6-7 kg. (The counter weight is about 5).

### 130. From Rowbotham, ND

This rather lengthy description comes down to stellar parallax. Stellar parallax is the observed difference in position due to Earth's motion through its orbit. By taking six months, as described by ND Rowbotham, you get the maximum parallax.

However, one has to appreciate that stars are bloody far off.

Which means you need extremely accurate readings - far more accurate than those by human eyes, which are rather bad. From Wikipedia, we get:

And there you have it; it has been observed. And therefore, we conclude that the earth moves. Wait, we knew that already, so it's confirmed again.Stellar parallax is so difficult to detect that its existence was the subject of much debate in astronomy for thousands of years. It was only first proven in 1838 when Friedrich Bessel made the first successful parallax measurement ever, for the star 61 Cygni, using a Fraunhofer heliometer at Königsberg Observatory.

### 131. Everything we've been told is a lie!

Apparently, NASA and modern astronomy (and ancient ones, too) are wrong.

A Neil deGrasse Tyson meme. Everyone loves Neil. |

The moon is

The

*clearly circular*and has in fact been proven to be*transparent*and*self-luminescent*. Of course, the fact that there is no physical mechanism known to mankind that could make it satisfy both of these doesn't oppose this claim, because it has been*proven*. That's why they refer to the proof... Wait, they don't?The

*fact*is that the moon has observed to be in Libration. You can find explanations (NASA) on youtube.The

*fact*is that the moon landing is real, and*can be independently verified*. That the Soviets accepted it should tell you a lot, if you have any idea of the political circumstances of the space race. Of course, you don't have to take their word for it. Even the TV series `The Big Bang Theory' has featured some evidence of the moon landing.### 132. Some unsubstantiated vagueness.

Composite Image of the Moon taken by the Galileo spacecraft, 7 december 1992. Via Wikipedia. |

To the right, we see an image of the moon. Note how it is not blueish.

There is a

*well known*effect called the Purkinje effect or dark adaption. The human eye is a measurement instrument which works within certain parameters. What this means is that it behaves differently, however slightly, under different conditions.For low illumination levels, the different behaviour is that it perceives blue far stronger than the other colours. Normally, it is a reasonable measurement of the spectrum of the light (how strong each monochromatic colour is). When illumination is low, it is not; it suppresses everything but blue. And that is why moon light is blueish/silver; it's white with slightly more blue than you normally perceive.

For high illumination levels, the different behaviour is that it perceives red stronger than the others. As a result, in daylight things seem more reddish, more golden.

There's an interesting figure called the Kruithof Curve. It's a bit complicated, so if you look into it you will also need colour temperature. It basically shows you colours on one axis, illumination on the other and perceived colour as an intensity-plot.

### 133. A weird reference to a non-pertinent journal.

I actually went after that reference, and found it. It's directly from Rowbotham, N.D., his book.

I'll directly address those. First, moonlight - even when concentrated over 6000 times - didn't heat anything. I'm not surprised. You shouldn't be either.

The full Moon is about 1,000,000 times fainter than the Sun. [Wikipedia/Moonlight]

The sun is at some $1000 \left[\frac{w}{m^2}\right]$. So, moonlight is at a thousand of that. A lens of 30 inches in diameter was used. We can calculate the surface area; the diameter is $76.2$ cm, the radius is $38.1$ cm, and the surface area is $\pi r^2 = \pi 0.381^2 \approx 0.456 \: [m^2]$. Therefore, the heating done by this concentrated moonlight is about $0.456 \cdot 6000 \cdot 0.001 \approx 2.74 W$.

Suppose that, for no apparent reason, the heated surface is a thermometer stuck in a glass of water (0.2 L, standard volume). How much would this water heat per second? Well, 0.2L of water is about 0.2kg of water, and it has heat capacity 4185.5 Joule per Kg per Kelvin. Therefore, it needs 837 J for one degree of heating. Using the concentrated lunar light above, that's about half a minute. This is very slow heating, and the heat would probably dissipate at a similar rate.

As a physicist, I'm going to state this clearly. Moonlight is just light; there's nothing special. It's made of photons, it broadly obeys the black body spectrum, and so on. There's nothing fancy.

### 134. Apparently, round things can't reflect.

Thank you google. Clearly, this spherical marble doesn't reflect light. At all. Nu-uh. |

Did you play with glass marbles as a child?

So they do reflect? Even though they are spherical?

### 135. You can see the blue sky through the moon!

Not really. The blue sky is an illusion because of rayleigh scattering in the atmosphere. This 'background' of blue light combines with the rays coming from the moon to give you that image.

### 136. Deny that the greek knew of the shape of the earth.

Basically, this entire argument is predicated on the idea of a spherical earth coming from Copernicus. But it didn't; copernicus was prosecuted because of heliocentricity, not because of a spherical earth. The difference being, it was accepted that the earth was round but people thought the sun went around it, and the earth is in the middle.

Copernicus saw no good reason for this, and correctly noted that the astronomical observations matched more closely to the sun in the middle, things turning about the sun, including the earth. Especially the orbits of the other planets make far more sense that way.

The current view in science is that both are viable. Einstein made it pretty clear; all views are equally acceptable.

I don't mind heliocentricity nor geocentricity. I do know the math becomes far simpler for the first compared to the latter. Much like I know the fourier domain allows many things to become rather simple, say, differentiation.

One of the more difficult parts of working in classical mechanics is choosing coordinates - point of views - that simplify the math.

I recall that, in the first year of the bachelor programme, a teacher had an optional lecture deriving kepler's laws. Because we had to work from Newton's law for a rotational system, this was an

*extremely*lengthy lecture. It took two hours, with some steps skipped, and about six blackboards.
As opposed to the video at the start; the derivation using lagrangian mechanics takes about half a page in Dr. Thijssen his Lecture notes.

### 137. Apparently, lunar eclipses aren't caused by what we know.

Wikipedia, diagram of earth's shadow. |

The explanation given is that, because both can be visible in the sky, this can't happen. As often, I disagree.

There is a short time where a mirage would be able to show the moon, from the penumbra, in part where it is still day. A mirage isn't even needed; there is just a small timespan near sunrise/sundown where your field of vision carries over the horizon, even more so if you're high up. In fact, it turns out to have happened in August, 1989.

So what about the claim that it happens at day, too? After googling a bit I came to another flat earth post, which noted:

The Pliny reference is straight from Rowbotham, N.D., again. It refers to an almanack where the same thing is said, not to Pliny.As early as the time of Pliny, there are records of lunar eclipses happening while both the Sun and Moon are visible in the sky. The Greenwich Royal Observatory recorded that “during the lunar eclipses of July 17th, 1590, November 3rd, 1648, June 16th, 1666, and May 26th, 1668 the moon rose eclipsed whilst the sun was still above the horizon.” McCulluch’s Geography recorded that “on September 20th, 1717 and April 20th, 1837 the moon appeared to rise eclipsed before the sun had set.” Sir Henry Holland also noted in his “Recollections of Past Life” the April 20th, 1837 phenomena where “the moon rose eclipsed before the sun set.” The Daily Telegraph recorded it happening again on January 17th, 1870, then again in July of the same year, and it continues to happen during lunar eclipses to this day.

I find the Greenwich Royal Observatory reference extremely interesting, as it was commissioned in 1675...

McCulluch is rather closer to the mark; it's just the effect of watching over the horizon when it is still day around you. The moon is dim, but it is still a lot brighter than most stars are.

Henry Holland, again, refers to the clipse before the sun sets. In fact, all of these are about the effect of the eclipse being visible just before the sun sets.

And it turns out that this is called selenelion, or horizontal eclipse. And it happens because of ... refraction in the atmosphere! It's a mirage!

### 138. No, the simple explanation is false; instead, have our arbitrary conjecture.

Calling something a "Law" doesn't make it true. The law of perspective is only ever used by flat earthers, and is only ever used when claiming that 'sinking below the horizon' isn't only explainable by a non-flat earth.

Of course, they have never formulated the law, demonstrated the law, done rigorous research into the law or whatsoever. It is rightly called the Conjecture of Perspective, with additional dodging the question built in.

Either way, we discussed this in #65.

### 139. Let's repeat #65.

And claim that a telescope resolves this.

Given that ship captains accepted the non-flat earth theory, and that they regularly carried telescopes around... Yeah, No.

### 140. Somehow, Physicists got the calculations for Foucault's Pendulum wrong.

Of course, how they're wrong isn't actually said; it is claimed that they are wrong. The talk about the initial conditions as if these are a sufficient explanation of why physicists got it wrong; but of course, the initial conditions and boundary conditions are part of the calculation.

There actually isn't a stationary pendulum anywhere on earth. You see, the earth rotates about its axis. Seriously though, this is silly. A pendulum is subject to several conservation laws; if nothing is there, then nothing is conserved. So you have to put in something, which it can then conserve.

### 141. Because copy/pasting isn't hard enough, here's a myth.

The coriolis/toilet myth isn't an argument against the spherical earth.

However, storm systems are - these are of sufficient scale for coriolis to be significant.

Cyclone refers to their cyclonic nature, with wind blowing counterclockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere. The opposite direction of circulation is due to the Coriolis effect. Wikipedia and Wikipedia.

### 142. Actually, sort of true. However..

So the explanation here isn't bad. Yes, images lose coherence over such a distance that they probably won't be visible over long distances.

However, things like 'pulsing' lights have far longer visibility, and can even be found in the 'light signal' of a photodiode if they aren't visible with the naked eye. I had to do this for an experiment once; it is one of the many techniques to evade strong (white) noise.

The explanation does certainly fail, however, when you notice that the horizon-distance is such that under normal circumstances, they obey the formula found using the non-flat earth theory. And, what's more, when you don't find it you do find non-normal circumstances...

Luckily, the point ends with a claim made without explanation or foundation.

### 143. A flat earth has a horizon?

Given that they don't explain what the horizon's supposed to be on their non-existing model, this is rather silly. The sun is a spherical gas cloud with nuclear hydrogen fusion in the core; it sends light in all directions. Realising this, it doesn't act as if it's a flash light hanging from the ceiling.

As for the final line; the answer is light scattering in the atmosphere and the angle of incident light changing as you pass the day/night line.

### 144. Ignore lunar libration, would you?

They forget to explain the sentence they opened with. Their non-existent model has the moon circling around, much like the sun; but that doesn't tell you why the southern hemisphere sees it vertically reversed from the northern hemisphere.

### 145. Let us depend on ignorance.

So here they basically explain what the non-existing model has to say about the moon. Apparently, it is a disc at a certain inclination. So, apparently the fact that 59% of the moon is visible due to libration has to be ignored.

### 146. Let us, once more, not understand simple explanations.

So the moon orbits the earth, with a full orbit completed roughly every 28 days. The earth, meanwhile, spins (causing day and night) which means the moon seems to orbit every day; that's just the view that everything spins around the earth instead.

Here, they show they don't understand that, claim the moon orbits the earth every day and that somehow, science got it wrong.

Science didn't; the moon apparently orbits the earth every day, due to a chance of reference frame (non-inertial). However, the moon actually orbits the earth every 28 days, which causes a shift in position for that perspective. The model becomes very complicated when you add that in.

Their libration-ignoring, vaguely mentioned moon is a disc that changes angle; so they have two separate motions (orbit and spin) instead of one motion (orbit) to explain the moon. And they still don't explain libration, which doesn't happen for the disc.

### 147. False claim (about a claim) means design.

The ' ball earth theory' doesn't claim an exact 400 times. In reality, the ratio of semi-major axes is about $389.17$. The ratio of equatorial radii is $400.66$; neither is

*precisely 500*.
Perhaps you shouldn't use a 0.2 arcminute accuracy instrument if you want to measure

*precisely*.### 148. Let's claim that something explained by a theory isn't in it.

Sidereal time versus solar time [Wikipedia]. |

So, let's talk about this. For many purposes, the stars are so far away that we can consider them a fixed background. Let's do so now.

With regards to the stars, so very far away, the distance the earth travels in its solar orbit isn't very significant.

However, for the solar orbit, it is important. Because of the displacement, it takes slightly longer for the solar rotation. Essentially, the way to measure this is to put a stick in the ground and note how long it takes for the sun and for a far away star to align.

For the sun, this takes about 24 hours; for the stars, it is about 4 minutes less. The first is called a solar day. The second is called a sidereal day. As you can imagine, because Earth's distance to the sun is not fixed, the sidereal day varies slightly throughout the year. The average length is called a stellar day.

### 149. Let's underestimate the distance to the stars.

So, similar to when we discussed stellar parallax, the clue is that the stars are really far away. So far away, in fact, that the motion of our planet around the sun barely matters, and the motion of our solar system only slightly more.

Enough to keep it constant? No. No, I agree, they should change.

Constellations throughout the ages. Via Halcyon Maps |

Ah, they do. Well, that makes sense.

For the record, "CE" means Common Era. Some people prefer BCE to BC, because the latter has a religious reference. AD is similar; it means Anno Domini, the year of our lord. Both are religious terms, and there are people that prefer non-religious annotations. For instance, consider a Judaist that has to write B.C. very often; what's up with that? (NB: Judaism is the religion; Jews are the ethnic group.)

### 150. Things close to polaris shouldn't look like they're close.

I'm not really sure about what point they're trying to make. Things that are close to polaris, where close is rather broad, will seem to move circles around it.

I'm guessing that they're forgetting about tilt, again.

Here is a pretty good explanation with pictures.

### 151. Let's reformulate something we already considered.

Earlier, we considered the Stellar Parallax. Because stars are so mindbogglingly far away, the stellar parallax is really, really small.

What they describe here, the movement of our solar system and our planet inside of it, would result in stellar parallax. We already know that, because of the distance involved, the parallax isn't really visible unless you use pretty precise instruments.

And that means that their claim that this leads to 'star-trail photos to be impossible' is bollocks. The parallax is too small to disturb the star-trails.

### 152. Flatter than a pancake.

We already know that local areas can seem flat; the flatter than a pancake article just shows this.

Does this somehow go against a non-flat earth? Of course not.

The data they used comes from a USGS Digital elevation Model. Because the Earth is not a perfect spheroid; it is an oblate spheroid, but geology is put on top of that - it can be that some local areas are flatter than you would expect.

What's more, the flatter than a pancake article shows that Kansas isn't even flat. If you look at the picture, you see that the pancake isn't a flat surface but that the edge is taken into account; that does raise its flatness a bit.

What's more, everything is flatter than a pancake. In a response to the flatter than a pancake article, a paper in geographical review states this:

In an article published as a spoof but based on actual data and legitimate algorithms, three geographers cleverly proved that “Kansas is flatter than a pancake” (Fonstad and others 2003). Their conclusion was widely reported by news media and accepted as proof by many people. The argument played well with a public already inclined to believe that Kansas is flat, but then-Director of the Kansas Geological Survey Lee Allison retorted that, by that measure, any state, even mountainous Colorado, would be flatter than a pancake. His point is readily comprehended if one imagines stretching a pancake to the size of a state. The pancake measured in the article was 130 millimeters, and its surface relief was 2 millimeters. Apply that ratio to the east-west dimension of Kansas, approximately 644 kilometers, and the state would need a mountain (2/130 x 664,000 meters) 9,908 meters tall in order not to be flatter than a pancake. Since the highest mountain in the world is 8,848 meters tall, every state in the U.S. is flatter than a pancake.

Surprise! Now, look also at the next point for why Kansas is so flat.

### 153. Let's make a lot of points we already made, in some form or another.

Things look really flat! Yes, they do, because the Earth is rather large compared to the distances you can see. Height, elevation, is defined from the (average) radius of the earth onwards, or compared to sea level or whatever.

Height is literally that difference; if the distance from the centre to the earth as measured is $r_H$, and the reference radius is $r$, then elevation/height is defined as $ h = r_H - r$. It's pretty simple.

So actually, the paper in #152 was evaluating the geography of the region; not whether or not it is flat as in 'flat earth'. Of course, the flat earthers didn't mention this, didn't they?

### 154. Let's try to distract them with something nearly true, then distract them with something false.

Same image, now with equal scales. |

As we discussed for the Applied Optics paper, lenses of cameras, glasses and the like do have some effect. But is that effect as big as they make it seem? Note that the scale of the two images they give you is very different. I used their images to scale the one on the top-right down to the scale of the one on the left, and then made sure they had equal width. The result is to the right.

As you can see, the effect of the lens isn't that much - it is, largely, due the larger field of vision.

The image on the left has a very small field of vision, due to the small window and the larger distance (1.5m?) of the camera from the window.

A quick calculation gives the field of vision as

$$\begin{align}

\theta &= \text{arctan}\left(\frac{0.3 \cdot 0.5}{1.5}} \\

&\approx 5.7 ^o\end{align}$$

An estimate at how far up you'd have to be to see any difference with the naked eye using my sagitta from #60 can be made. From the numbers the flat earthers give us, baumgartner was about $40$ km up.

Because the formula is $\propto \sqrt{h}$, the change flattens out. Anyway, the result can be seen in the following plot, which explains why the image on the right doesn't see it. The image on the left, on the other hand, has a far, far larger view angle. For that view angle, we find a different plot... I estimated the left view angle at $\theta=40$. It is the one to the right.

### 155. The horizon remains a mystery.

Once more, they explain how they don't understand what the horizon is.

Well, that was a short response :)

### 156. Clearly, it wasn't called `200 unique reasons'.

Another one about camera lenses. Once more, I must point out that there certainly are amateur photographs of the curvature. As you can see in the images on SpacePhoto's Tumblr, the curvature there is visible (at ~22.5 miles - certainly could be visible, according to our calculations). The team doesn't detail it, but their description seems certainly educated enough to have used a lens corrector.

Another excellent example is a amateur built rocket with a non-"fish eye" lens.

### 157. Let's repeat another one.

Remember how we went over the atmosphere turning with the earth, and airplane velocities being measured relatively to a point on the earth's surface, usually a control tower?

Yes, they are using that one again. You'll also remember how I backed the parts about airplanes up with information taken from books on airplane engineering.

Yes, they are using that one again. You'll also remember how I backed the parts about airplanes up with information taken from books on airplane engineering.

### 158. Apparently, rotational motion is hard, too.

The atmosphere is spinning at a constant rate. It's not exactly true, but for the rain and fireworks it will do.

The claim here is that the different rates of spinning atmosphere for these things dropping down would give different motion.

There are two points to make here. First, there certainly is an effect due to the earth spinning:

Second, have you never noticed that rain drops at an angle, even if there is no or little wind?

As you can see, it's pretty straight.

So, the real question is: Why would you launch a rocket at an angle? The answer is quite simple; because of winds. Firing at an angle is usually done because you want it to land somewhere, or leave the atmosphere somewhere. I can also speculate that going with the wind (on average) gives a more predictable trajectory.

Apparently, the flat earthers think it is more likely that the government is hiding hologram technology.

That wasn't so hard, was it?

Wikipedia tells us where the satellites are:

I think I did refer to this one before.This one is also quite good.Hooke, following a 1679 suggestion from Newton, tried unsuccessfully to verify the predicted eastward deviation of a body dropped from a height of 8.2 meters, but definitive results were only obtained later, in the late 18th and early 19th century, by Giovanni Battista Guglielmini in Bologna, Johann Friedrich Benzenberg in Hamburg and Ferdinand Reich in Freiberg, using taller towers and carefully released weights.[n 1] A ball dropped from a height of 158.5 m (520 ft) departed by 27.4 mm (1.08 in) from the vertical compared with a calculated value of 28.1 mm (1.11 in). [Wikipedia]

Second, have you never noticed that rain drops at an angle, even if there is no or little wind?

### 159. Apparently, they think things are sudden.

The simple truth of the matter is that the earth becomes thinner and thinner as you move further outwards. Additionally, space isn't a vacuum - it is just close to a vacuum.

### 160. Physics, still evasive.

Dear Flat-earthers, please, please join a high school if you think there's any sense in this argument.

Rockets and planes aren't propulsed by the air pushing on it. Yes, they generate lift to offset gravity, or at least, planes do. Rockets really don't see the point.

You see, an elementary calculation leads to the escape velocity, the velocity required for an object (pointing 'up' or outwards only) to escape the gravitational field. This has to be corrected by an estimate of the energy lost due to drag. A good guess can be made by numerical approximations; yet it becomes clearer still when standard atmospheric data is used to simulate the trajectory using a computer.

Either way, Newton's first law will keep the rocket going. That's all that's needed; sufficient velocity to beat the drag and gravity.

And why would it start spinning? There's no reason for that - no force that acts such as to cause spinning.

Note that many rockets do spin; this is merely because spinning stabilises the rocket, reducing the influence of wind and such.

### 161. Spherical orbits are hard.

This again (#15)? We've already discussed this one, too. Airplanes constantly do correct downwards; the force that acts to do this is called gravity.

### 162. Rocket's are fired sensibly, therefore, they don't happen.

So, let's first watch this video of the stratos II+ rocket launch:

So, the real question is: Why would you launch a rocket at an angle? The answer is quite simple; because of winds. Firing at an angle is usually done because you want it to land somewhere, or leave the atmosphere somewhere. I can also speculate that going with the wind (on average) gives a more predictable trajectory.

### 163. Libel against NASA always works.

Just a lot of claims without any back up.

The closest thing (quora) I can find is that some videos, which people think are in space, are actually the underwater training facilities. (Note that there are good reasons to train underwater).

### 164. Non-cited analysis shows that our claims our true, period.

There is but one proper response.

I like this meme. Feel free to tell me whom is actually in it; I don't know. |

### 165. I can see it, therefore it isn't real.

The 'documentary' he refers too seems too much of a waste of time. As a number of acquaintances say, I will watch your video for a large amount of money. I'd much rather spend my time learning more about gravity, using Sean Carroll's excellent book.

So here is a video of the ISS through a telescope. It seems to be the one the picture they use come from, but I can't be sure. Real objects often look similar through a telescope. Here is another one. There's a lot of people trying to do this sort of stuff.

Apparently, the flat earthers think it is more likely that the government is hiding hologram technology.

As for why the sky behind it usually seems so dark; it's because of zooming in. There's not that many stars visible with a simple telescope, and the ISS is relatively bright (because it is close and reflects sunlight [ref]).

The real question is mostly why the discolouration happens. I don't know, and I can't really be bothered to find out. It's probably something to do with interference and the telescope used. It could even be something as simple as a halo due to some water vapour.

### 166. Let's tag on the conspiracies.

Freemasons are becoming a regular feature now. I'd like to make on thing clear about the Illuminati: We do not, in fact, exist.

Geostationary orbits aren't that weird; they're just a result from some elementary, classical calculations.

LORAN wasn't very accurate; accuracy up to tens of miles. DECCA was from a few meters up to a mile, depending on various conditions. GPS, using a correction based on Einstein's theory of gravity (general relativity), offers accuracies of a steady 5-10 metres. Europe's Galileo will go for one metre public, and

*one centimetre*for encrypted (commercial) use.
That's right, a centimetre.

GPS is satellite based, by the way - I'm not sure why they claim it isn't.

### 167. Temperature and heat transfer aren't the same.

The story here is that while the gas can be at high temperature, it exchanges nearly no heat with the satellites and regular radiation makes sure the satellites lose it rather quickly.

The highly diluted gas in this layer can reach 2,500 °C (4,530 °F) during the day. Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat. A normal thermometer would be significantly below 0 °C (32 °F), because the energy lost by thermal radiation would exceed the energy acquired from the atmospheric gas by direct contact. In the anacoustic zone above 160 kilometres (99 mi), the density is so low that molecular interactions are too infrequent to permit the transmission of sound. [Wikipedia].

That wasn't so hard, was it?

### 168. Satellite phones and low-tech countries.

A satellite phone basically uses a satellite, instead of a cell phone tower, for its communications. Communication satellites are geostationary, which is what they wanted to deny in #166.

Have they considered that there's less satellites in range of these countries?

Wikipedia does explain why the range isn't optimal.

### 169. Angle of TV dishes.

This page explains how the dish picks up signals.

Wikipedia tells us where the satellites are:

Satellites used for television signals are generally in either naturally highly elliptical (with inclination of +/−63.4 degrees and orbital period of about twelve hours, also known as Molniya orbit) or geostationary orbit 37,000 km (23,000 mi) above the earth's equator.

The Molniya orbits are mostly high up (elevation) as far as I can see. But this explains why you don't point the dish straight up; there's usually nothing there.

### 170. People claim things.

That's not an argument. That people claim to see satellites and aren't correct doesn't mean they aren't up there.

Note that you can see them; in lower orbits, the bigger ones are very bright compared to the night sky. This is simply because of the distances involved.

### 171. Distance is hard, too.

Basically, the argument here is that we have lots of pictures so a direct image of the full ball earth should be available. Oh, and all composites are CGI's. In a sense, they are; a computer makes the composition that is the full picture.

In another sense, they're the real deal. Most satellites are rather close to the earth, as NASA explains. Let's assume for a moment that all are at heights $h < 35780$ km. From our horizon estimation, $d=3.86\sqrt{h}$, we know that an object in orbit at $\left(\frac{6731}{3.86}\right)^2\approx 3000$km height can see the entire earth (horizon distance is earth radius).

Using that estimate, the ISS at about $350$ km (estimate taken from orbit graph, Wikipedia) would see about 34% of the Earth's surface. The ISS has a live stream, but I wanted to show you a nice video of it. If it streams badly, set it to somewhat lower quality (it goes up to 4k).

Before we start, it is clearly not a fisheye lens; you can see flat surfaces (solar panels?) in the sides of the video, and these are clearly flat.

Our distance estimation seems to work; around 42, they are a bit further looking downward (i.e. the field of vision is good to estimate the distance between their 'surface position' and the horizon) and about 30% seems right for what they can see in total. Just a rough estimate, mind you.

Here's some astronauts with a video from the ISS. NASA has a lot of material on there, because the project is awesome.

### 172. Let's not look closely and ignore accuracy.

The ISS has an orbital period of about one and a half hour. Given the time it has to look at one piece of the sky, I don't find it weird that most movement isn't visible.

The Galileo time-lapse (from 1990!) does show moving cloud fields, but you have to look closely. Remember, those are global weather patterns, not local ones; the local ones are not resolvable by the cameras onboard of that spacecraft a quarter century ago.

### 173. Let's be unfounded, like always.

4 Images of the Netherlands by the ISS. Those clouds sure look the same. |

Present your solid proof, seriously. Don't say it is there; show it. Their image shows a piece of blue and white which is open in an image editor. Now, I'm impressed - I wouldn't have expected them to actually show the image editor.

You can search for NASA's photographs. I choose 'north sea' in the Netherlands. I also put together a collage. These are ISS images, so not of the full ball earth. In these images it is very, very clear that the cloud patterns are very much not the same. Of course, with today its technology you can generate clouds; it is one of the reasons we're able to have climate change models.

And now, for the grand finale; from Nasa's image archives again, this time looking for blue marble. These are two images showing the americas and, most notably, one of them has a distinguishable feature near Mexico and California.

These images show how the clouds are exactly the same. |

### 174. Pareidolia is fun, too.

Pareidolia is the psychological phenomenon where you see patterns that aren't there.

They claim that Disney's pluto is in the background of the 2015 Pluto image. Nasa's press doesn't say anything about it, and the LORRI gallery doesn't show any, either. Apparently it went viral on twitter, but it needs a lot of imagination.

### 175. Unnamed professionals, in a report in an unknown peer-reviewed source.

Apparently, `professionals` have dissected NASA images and found `undeniable proof' of computer editing.

Neil thinks they should cite, too. |

So, which specific professionals? Where did they report their findings? Was it peer-reviewed? Of what images?

### 176. We forgot the thing rotates around its axis.

Apparently, they did. They show an image where the relative size of the continents is drastically different! No! Outrageous!

And the colourations of different cameras are different too! No!

Okay, I'm not even going to take this seriously. These are images of the Notre dame. They're at different angles and in different colours. Therefore, it doesn't exist.

### 177. We hadn't had a good conspiracy for a bit.

Reference to a documentary, too. Well, it's not exactly a solid thing.

It's not as if you can verify the moon landing without listening to NASA. Wait, is that a link? Oh, I guess you can.

### 178. `People' again.

Which people? Where do they claim this?

And there'd be a few differences because of map projections; a sphere can't be projected onto a plane while keeping the distances similar.

And there'd be a few differences because of map projections; a sphere can't be projected onto a plane while keeping the distances similar.

### 179. Airplanes again.

Airplane flights going counterclockwise and clockwise should have different travel times. We've already discussed this. The answer is: drag.

The slightly longer answer is; the atmosphere is dragged along with the earth, and drags the plane with it.

### 180. Because we haven't just said this.

The same argument, again.

### 181. A novel argument.

Nope. Same as #179, #180.

### 182. And again.

### 183. Had we done airplanes flying yet?

### 184. I think we have.

### 185. You can feel a car moving, therefore bollocks.

Yes, you can feel a car moving because of there being bumps in the road, and similar things.

Without that, you couldn't. Compare the situation to an airplane without turbulence; you can only feel it changing, accelerating.

As for the apparent force of you moving around in a semi-circular orbit, it's called gravity.

### 186. Let's not understand motion sickness, too.

Motion sickness is because of a difference between perceived and measured movement. As gravity provides the 'rotating' acceleration, the perceived and measured movement are the same. This stuff isn't very hard.

### 187. Earth experiences drag, for some reason.

The earth spins in space; the drag is so far near zero that it could be a list of citations by a flat earther.

It is claimed that this was never measured; yet...

The Earth rotates once in about 24 hours with respect to the sun and once every 23 hours, 56 minutes and 4 seconds with respect to the stars (see below). Earth's rotation is slowing slightly with time; thus, a day was shorter in the past. This is due to the tidal effects the Moon has on Earth's rotation. Atomic clocks show that a modern-day is longer by about 1.7 milliseconds than a century ago, slowly increasing the rate at which UTC is adjusted by leap seconds.

Atomic clocks for the win.

### 188. Let's be dishonest.

The oblate spheroid was predicted by Newton, and has been known since before NASA was formed.

It is slightly pear-shaped. To first order, the flattening comes in. To the next order, other terms come in. Most significantly is that the upper hemisphere has a slightly smaller radius than the lower hemisphere. And therefore, it is slightly pear-shaped.

The shape is pretty clear, but it's only 1/300 of the radius smaller at the poles. So yes, it looks like a spheroid.

### 189. Religious books.

So? There's also a story that a turtle, carrying the world on its back, crawled from the primeval soup. The Aztec believed a man's beating heart had to be ripped from his chest, every morning, to make the sun rise (I don't know how true that is, but human sacrifice did feature there).

There's religious people out there that think the lot of us should die. At the moment of writing, the world is in uproar because of several religious terrorist attacks (November 2015 Paris attacks and Beirut attacks), mostly the first and the second as a fashion statement.

And you think that there is such a thing as 'scriptural proof'? That it has any credibility beyond the church?

### 190. Ancient people though the earth was flat.

They also would've considered modern technology to be magic. So? How is this relevant?

It was hardly a minority view; it was the view of those learned, of sailors and such like. Also, argumentum ad populum; just because many people believe it, it doesn't have to be true.

### 191. Freemasons!

Ohnoes, freemasons! Well, that settles it.

### 192. A quote from an old book, which doesn't cite.

A book from 1901 is quoted, and in the quote it claims things. So? The book itself, of course, is already a flat earth book.

### 193. A child's understanding.

That's the argument; no child would come up with this by himself or herself.

Children don't regularly travel about, believe in santa claus, and usually don't look about enough to figure out science, physics or what not.

I'm not sure why a Child's perception would be an argument. Are you?

### 194. A dead author didn't physics.

This argument explains how the author from #192 didn't know physics. All right, good argument.

### 195. Magnetism of Gravity?

What is up with that? One of the great challenges of modern physics is to unite gravity and electromagnetism (or more specifically, the standard model). How can anyone that had the least bit of education confuse the two?

Also, the Cavendish experiment is a practical example on the smallest scale.

### 196. An arbitrary person denied things, too.

Some arbitrary person also denied everything. Each and every point in that quotation has been considered in this blog.

### 197. Not assuming you're special is wrong, somehow.

Apparently, not having megalomania is a bad thing.

This entire argument is basically that the motive for the extra-massive conspiracy is that the megalomania of some humans is challenged, leading people away from what appears to be the Christian religion.

Oh, and they make a nice demonstration that they haven't any morality, too:

*"then all that really matters is me, me, me."*Morality is not exclusively related to religion; some systems of morality can come from religion, certainly, but it is not the only source. Aretaic ethics, Consequentialism and duty ethics are three big categories of moral frameworks defined by philosophers.
It has, for a while, been clear that the most religious nations aren't the most moral ones; and that the most secular ones often are. As the religious often throw this at atheists, the response I most commonly see is "

*Scandinavia. Your point is now invalid."*And indeed, some of the most secular countries in the world, countries that had less to do with the Holy Roman Empire, are among the highest in social welfare, happiness and morality.### 198. More freemasons and Illuminati.

I want to make one thing clear about the Illuminati: We do not, in fact, exist.

### 199. Someone else that denied.

Okay.

### 200. Finally, from Rowbotham, N.D.

Rowbotham, N.D. concludes that his misunderstanding, outright dishonesty and so forth lead to an inconsistent system.

### Conclusion.

That was a very, very long post. I can barely fathom how long it became.

Working on this, I realised that when the conspiracists figure out I've rebuked the entire article, they're likely to delete and reshuffle. For that purpose, I've backed up the entire article they wrote. It's over here, daimonie.com/Archive/200, and will remain there.

I invite you to think long and hard about this; why people would want to delude themselves, but also, how easy it is for gish gallop to sound plausible.

To finish, I'll repeat my meme:

My Meme, again. This is likely the only place you'll ever find it. |