Saturday, April 30, 2016

Flat Earth Misconceptions

I'm so tired of repeating myself to Flat-Earthers. It's like they revel in doing their math wrong, but there's not enough characters in twitter to send them the proper equations. So I'll address a few classic pics here to explain how it's done.

The distant mountain picture

This is a favorite among flat-earthers, and it's pretty easy to show they get the math wrong.  Someone told this guy that he can use a simple linear fit for earth's CURVED surface (8 inches per mile or some such tripe). I'm not sure where this approximation came from, but it obviously doesn't fit a curve very well. Anyhow, here's an example:

Here, we have the classic flat-earther example of a "mountain that's too far away to see." First, let's check the facts. After a bit of digging, I figured out that they're claiming this is Mt. Denali.  That's 140 miles away, with an elevation of roughly 20,000 ft.

It's actually 700 ft at the summit of the Hilltop Ski Resort, but it turns out that won't matter in the end.  
Anyhow, let's assume they're right and it's Denali.  First, calculate the distance and angle to the horizon from an elevation of 700 ft.  I'll let WikiHow explain how to do it. I'm using the arccos formula: 

$ d = r  \arccos(\frac{r}{r+h}) $

The horizon from Hilltop is calculated here at 32 miles.
The horizon from Denali is calculated here at 174 miles.

Suppose the picture were taken from the parking lot of Hilltop instead of the top.  The horizon is still 27.5 miles away, well within the range of Denali's summit. 

32 + 174 means you can see the tip of Denali from as far as 206 mi at an altitude of 700 ft (assuming there's no mountains in between and neglecting atmospheric lensing, of course). This is obviously not a problem for observation from 140 miles away.

Just for convenience, we could figure out how tall an object needs to be to see it from a distance, x. We simply add the horizon distance to the solution to our first equation for h:

$ \cos( \frac{d}{r} ) = \frac{r}{r+h} $

$ h = \frac{r}{cos(\frac{d}{r})} - r $

d = 140-32 = 108
h = 7700 ft tall. 

This means we should be able to see the top 12,300 of Denali. 

Time Zone Solar Models

This is another flat earther hand-waving explanation for time-zones.  

Even the most casual thought regarding this picture and you'll realize that since the sun is above a flat earth, you'll be able to see it at all times from any location on earth. Turn this model on its side and draw a vector from any point on earth to the sun and you'll see that it never sets.  The figure below shows this and explains the problems.

If there's some other way to interpret the animation above, I sure can't think of it.  Send me a comment or tweet to help me understand.
I'm left wondering if flat earthers lack all spatial reasoning. It sure seems that way.

Pilots Would End up in Space!

This one is so bizarre that it takes a little work to wrap your head around.  Essentially, the argument is that airline pilots would need to adjust their elevation by 1 degree every 6 or 7 minutes to keep from climbing higher and higher and crash into the firmament or flying into space.  There's some specious assumptions here:
  • That airplanes fly along laser-like straight lines
  • Airplanes will continue climbing at the same rate for a given angle of attack even as the air thins.
  • That the pilot has the sensitive equipment to recognize one-degree drift over 6 minutes
In the same way as you might make minor course corrections along a straight road to keep your car centered in your lane, the autopilot (or the real pilot) is constantly adjusting to maintain a reasonably constant altitude and heading.  These corrections are happening at a rate of tens to hundreds per minute.  To the pilot or autopilot, this feels like maintaining altitude.  In reality, it's conforming the flight to the curvature of earth.  This publication described the typical roughness of a commercial flight.  I've pulled out the "rough cruise" section because the constant readjustment of altitude is more clearly apparent.  Same thing happens in "smooth cruise", but more gradually.

Simple computer models can help us understand when and where we'll be able to observe curvature of a sphere.  The key factors which affect the appearance of curvature are:

  • Altitude above the surface
  • Camera Focal Length (or field of view)
  • Radius of the sphere
I'm sure someone could write an expression  for the apparent curvature versus these parameters, but it's easier to just show you. I'm using an open-source tool called Celestia which accurately presents the position, sizes, and velocities of celestial objects.  I highly recommend it. It's fun to play with.

Changing the Field of View

Here's the  same view as the Field of View is altered.  This is like zooming out on your camera.  The FOV is reported in the bottom-right while the location (constant) is presented in the upper-left.

Changing the Altitude

This one is a little more obvious.  As you move to higher and higher altitudes, the curvature becomes more apparent.
Note that the Distance is 30 km. That's a decimal place, not a comma.

Radius Matters Too

This one is just for fun.  Here's some spheres of different radii from the same distance.

Mythbusters are shills?

Well, as we've shown, the curvature isn't expected to be visible at low altitudes. Here's Adam Savage  at a 12 mile altitude witnessing the curvature for himself.  So I suppose he's got to be a liar now, eh?

The Moon / Spinning Earth can't be felt!

Moon: The gravitational acceleration of the moon on the surface of earth is given by $ a = G m_{moon}/r^2 $ or roughly $ 3.6 \times 10^{-5}  m/s^2  $ compared to the $ 9.8  m/s^2 $ I measured in high school for earth's gravity.  That's less than one part in a million.  A hard thing to measure.

Rotation: The acceleration of an object on the equator due to circular motion from the rotation of the earth is given by: $ a = \omega^2 / r $.  This also tiny at $ 0.034 m/s^2 $. This effect (0.34%) MIGHT be measurable by exceptionally sensitive equipment and a skilled scientist, but these are the sorts of people the Flat Earth crowd seems to consider untrustworthy.
Edit: I fixed my math above. Rotation is actually much more important than I originally calculated. Thanks to @TheOlifant for catching my error:


The mentality of flat-earthers seems to be very similar to that of anti-vaxers and deeply religious.  The believer thinks they've figured out that most of humanity is wrong, and that their answer is the right answer. They often tell you to "research it," and couple commands with insults "stupid" or "dummy" or "sheep." 
These believers think they've figured out what "they" don't want you to know.  The "they" varies between people, but it seems to be illuminati, the government, or the Free Masons. For devout Christians or Muslims, the "they" is Satan, heretics, or demons.  
These believers pride themselves in being different. They think they're visionaries for knowing the truth when everyone else has it wrong.  Despite having no formal training in the specific scientific claims they reject, they feel sure that all the professional scientists have been deceived by the "they."
What's particularly interesting is that these people seem to blindly follow (IMO obvious) quacks. Some guy with a YouTube channel is seen as more reliable than all the world's scientists. They wave off these brilliant scientists by presuming they've never actually TESTED any of the claims they learned in science text books without seeming to notice that:

  1. The YouTube quack has never tested his flat earth claims. At best their "evidence" seems to be that they find actual physics hard to understand or inconsistent with scripture.
  2. Scientists actually do verify the basics. They build more complex experiments on top of them, so if the basics weren't right, nothing would work.

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