How Al-Biruni Calculated the Circumference of the Earth Using a Mountain in the 11th Century

Can we know the shape of the Earth?

or

How Al-Biruni Calculated the Circumference of the Earth Using a Mountain in the 11th century

As you might recall from my earlier posts, I've used the Sun at local Solar Noon (when the Sun is at its zenith) on the Equinox as a way to know the shape of the Earth.  But all we need is some point that is high enough to be seen at some significant distance and Mountain peaks are much easier to experiment with than a moving Sun.

If the ground is a flat plane then we can simply triangulate using the mountain peak at different distances and get the SAME answer for the height using simple trigonometry:

For example, if the ground was flat, we could find h by simply measuring our angle α at two different distances, where Δd is the distance between the two observation points using the formula:

    Δd * tan(α₀) * tan(α₁)
h = ----------------------
      tan(α₁) - tan(α₀)

Can you prove that?

This seems complex but lets think about what the tangent function REALLY does. You put in an ANGLE and you get back the matching SLOPE.  And SLOPE is simply RISE over RUN (rise/run). That's why we have the relationship: tan(θ) = opposite/adjacent -- once you understand that it is simply converting an angle into a slope I think things get so much simpler.  All we are really doing here is a unit conversion.

And since it is rise/run we can expand this equation by replacing the tangent function with the equation for each of our slopes and we get:

(d₀-d₁) * (h/d₀) * (h/d₁)
-------------------------
        h/d₁ - h/d₀

Now what we want to do is prove (mathematically) that this equation would give us the value for h if we plugged in all the values.  To do that we need to show that we can simplify the equation and everything will cancel out, leaving us the value for h at the end.

First, let's multiply out our numerator and move the denominator over to keep things easier to read (dividing by something is the same as multiplying by the inverse):

 h*h*(d₀-d₁)          1

------------- * ------------

   d₀ * d₁       h/d₁ - h/d₀

Next, we can join the denominator on the left by multiplying both fractions by the Least Common Denominator (which is [d₀*d₁], and simplify our new denominator):

 h*h*(d₀-d₁)      d₀ * d₁

------------- * ----------

   d₀ * d₁       h*(d₀-d₁)

Now we can see that [d₀ * d₁] cancels out leaving:

 h*h*(d₀-d₁)

-------------

 h*(d₀-d₁)

And we can cancel out the common term [h*(d₀-d₁)] from the numerator and denominator which leaves us with just:

    h

[Or let Wolfram|Alpha do it]

So we have proven mathematically that our starting (slightly complex) equation will give us the value we need. Our problem is we don't know the value of h to find h/d₀ directly but since we do know what the value of h/d₀ is (because we have the angle and h/d₀=tan(α₀)) we can plug in the values and get the answer back out of the equation.  So we don't even need the distance to the mountain in this case, just the distance between our two observations and the angle to the peak is sufficient to calculate the height AS LONG AS our ground plane is flat.

However, when you try this over longer distances, you find that it doesn't quite work. As you get further away from the mountain the mountain appears to shrink. The further away you get the more it shrinks.  If you try it with a streetlamp you'll find that it works very well because the ground is flatter than your measurement error over short distances.  But with mountains or tall buildings you begin to measure the curvature of the Earth appearing to make the object shorter and shorter with distance.

And the amount it shrinks is pretty consistent no matter which direction you go from the mountain so we know the curvature is pretty even in all directions and at all points around the Earth where we try to measure tall objects.

It is because of this, almost** equal shrinkage in all directions, at all points around the Earth, that we can conclude the Earth is a spheroid (and with better measurement accuracy we can tell it isn't EXACTLY equal in all directions but we can tell that it's a little bit oblate or squished making it a little bit wider around the equator than around the poles).

** On a sphere it would be equal, on our slightly oblate/squished ellipsoid Earth it varies slightly, mainly with latitude.


So what we find is that our ground is NOT flat because our angles keep changing with distance as the ground is rotating underneath us, throwing off our calculation:

This doesn't mean that triangulation is useless (but it fairly conclusively, confirms the shape of the Earth as being curved).  But what we need to be able to do is take our measurements from close enough to the mountain and from some height so we can still see the same base point on the mountain so that our baseline is flat, based on a direct line-of-sight.

Ok, so we can use this to calculate the height of the mountain as long as we make sure each observation point measures the angle between the same spot at the base of the mountain and the peak AND we now know that the Earth is curved.

What Al-Biruni figured out in the 11th century was that if he could find the height of the mountain peak as above (using angles and distances) AND find the angle between a level on the peak and the horizon (shown as angle β below) then this would also be the angle made at the center of the circle, and from this he could estimate the Radius of the Earth, and with the Radius of the Earth it is easy to find the circumference (r·2π).

Here is the geometry that Al-Biruni had in mind, at point A at the peak of the mountain we measure the angle of the dip to the horizon (greatly exaggerated here, it will usually be a smaller angle):

Instead of a tangent, this time we're going to look at the sin function.

What we know about sin is that it is also a ratio, just like our tangent function, consider a right triangle:

sin(x) = opposite / hypotenuse

What 'opposite' means here is the length of the side opposite the angle in question - so we have the following relationships:

sin(A) = a/c

sin(B) = b/c

sin(C) = c/c

From this it is easy to show that the Law of Sines holds:

  a        b       c

------ = ----- = ------

sin(A)   sin(B)  sin(C)

By simply expanding our formula, for example:

 b    b c

--- = -*- = c

b/c   1 b

We also know that the sum of all the angles must equal 180° and that one of our angles is 90° so when we measure the angle β we also can know that α is simply 90°-β and we also know that sin(90°-β) = cos(β) [I'll leave this as an exercise for the reader].

Now armed with these identities we can write out the equations for this diagram of the Earth using our Law of Sines:

  b       c

----- = ------

sin(B)  sin(C)

And we substitute in our known values, where:

b = R

sin(B) = sin(90°-β) = cos(β)

c = R+h 

sin(C) = sin(90°) = 1

So that gives us a starting equation where we can solve for R.

  R         

------- = R + h

 cos(β)

Isolate R to one side by subtracting R(cos(β)/cos(β)) from both sides, you will see why we added the (cos(β)/cos(β)) in the next step:

   R      R cos(β)           R cos(β)

------- - -------- = R - h - --------

 cos(β)    cos(β)             cos(β)

Which simplifies to:

     1

R(------ - 1) = h

  cos(β)

We can then multiply both sides by the inverse, or 1/(1/cos(β)-1) rewritten as cos(β)/(1-cos(β)) which gives us:

    h cos(β)

R = --------

    1-cos(β)

Which can also be written:

       h
R = -------- = h/(sec(β)-1)
    sec(β)-1

So now we have the portion of our triangle represented by R in terms of the height of our mountain and the angle between level at the top of the mountain and a distant point on the horizon.

Let's say we're up on a Mountain at 2000 meters.  The dip angle to the horizon should be about 1.43546°.   Plugging that into our formula:

2000/(sec(1.43546°)-1) = 6.371×10⁶

What if we're up at 33,709 feet?

We would expect this angle to be about 3.25° - and since the iPhone Field of View is about 63.5° and that image is about 1077 pixels high we expect that to be about 55 pixels high (this is not exactly true because the lens is not perfectly linear, but it's a small section near the center so this is a good approximation).  PLEASE NOTE: The angle in question here is the Elevation Angle - the "Horizon Angle" is how much the horizon is tilted clockwise/counterclockwise.

What do you think, does that look like about where the horizon is?  It's hard to tell because of the clouds and atmospheric distortion in this image but hey, it's what I have (the horizon most definitely does not begin at the white section).

Next time you are going on a flight grab a Theodolite app and maybe you'll get a better view of the horizon and you can estimate the circumference of the Earth.  Can you beat Al-Biruni's value which was correct to within ~200 miles?

Post your results in the comments section!

I have several flights coming up this month so I'll try to get some better data to share.

Flat Earth Follies: GoFast July 14, 2014 Rocket Launch

Flat Earth Claim

The Earth is Flat! 100% Video Proof. Globe Earth Debunked

GoFast July 14, 2014 Rocket Launch Proves the Earth is Flat because you can see the moon.

Figure 1. The Claim

Also here and Reffi made a video about it.

Analysis

Prediction - this will consist of more (hopefully well-intentioned) Flat Earth misrepresentations, misunderstandings, and conclusion jumping, as evidenced in the title of Reffi's video:

The Earth is Flat! 100% Video Proof. Globe Earth Debunked

This implies they have the necessary skills and have taken the necessary care in evaluating the evidence and I think it becomes evident that they do not have the necessary skills and have not taken the necessary care in evaluating the evidence here.  I hope I have done a better job, but it is foolish to call ANY analysis "100% Proof" of anything.  So I simply offer my refutation with as much detail as I can.

The Scene

Somewhere in the Black Rock Desert, Nevada
Morning July 14, 2014
14:32 UTC / 07:32AM Pacific 

Figure 2. Location

Ky Michaelson and the Civilian Space eXploration Team (CSXT) launch the fastest amateur rocket ever launched.  In well under two minutes from launch it would hit 3,580 mph and reach 385,800 feet above mean sea level (about 73.1 miles or 117.59km) before falling back to Earth.

In the posted CSXT video, the moon is first visible approximately 1 min 22 seconds into launch, very near apex, so that puts us at roughly 14:34 UTC time (a few minutes either side of the exact time does not make a perceptible change in the astronomical positions) and at roughly 117.59km above the Earth.

I do not know their exact GPS launch coordinates, nor exactly where the rocket was at the time of observation but we know the position well enough and we know the times and altitudes well enough that we can check the view from that area and see if it comports with the Globe model or the Flat Earth model.

Astronomical Data

Black Rock Desert
Latitude: N 40°56'10.82'' / 40.93634°
Longitude: W 119°5'30.48'' / -119.09180°
Ground Elevation: 1190m
Rocket Altitude: 385,800' / 73.1 miles / 117.59 km

Lunar Position from Black Rock Desert (mooncalc.org) at 07:34 UTC-7 (localtime)
Altitude: 8.83°
Azimuth: 248.27°

Figure 3. Moon position from Black Rock Desert

So the moon would have appeared 8.83° over the horizon at this point from the ground view.  But our rocket is 73.1 miles up so we get to see a little more room between the Moon and our Horizon (about 10.9° between level and where the Horizon falls ~764.13 miles away from our rocket).

I also found the sub-lunar point at the time of observation, which is the point on Earth where the moon would have appeared directly overhead.

Sub-lunar Point
Latitude: S 9°44'0.31'' / -9.73342°
Longitude: E 172°27'38.31'' / 172.46064°

Figure 4. Sub-lunar point at time of observation

Now the claim was that the moon was "over Australia" but this is actually some 1700 miles off the coast at an azimuth/direction of about 65.5° (North-East), or due north of New Zealand.

You can see in this timeanddate.com map that the moonlight is easily reaching western Nevada from this point.

So I'm not sure even on what basis these people then claim that you couldn't see the moon at this time "on a Globe model".  But, please, tell me exactly where the moon "should be" on the Globe model to result in this view - because based on my analysis here it is exactly where it should be.

Celestia

Now, let's use Celestia to see if this view comports with our Globe model.

For this exercise I urge the read to review my notes on Curvilinear (fish-eye) lens distortion in my post on High Altitude observations.  But for a very quick review... if you took a perfect grid of lines and took a picture of it with a GoPro (which has a curvilinear lens) it would look like this, with the lines bowed out from the middle:

Figure 5. Effect of Curvilinear lens distortion on a grid

This distortion is VERY easy to correct, you just have to bend the lines back to straight which kinds of compresses the view in the middle a bit.  To make a long story short, the fact that they used a GoPro will not present only the slightest difficulty to our evaluation in that the image we will see in Celestia will be perfectly rectilinear and we will have to apply some correction to the GoPro image to compare them.  Maybe the Celestia team will add curvilinear projection at some point.

Once you have Celestia (which is a free download) running here is how to setup your view to (roughly) match that of the rocket.  I like to turn on orbits and a grid for better orientation, you can set it up however you wish and then proceed.

The first thing we want to do is stop the timer.  Hit [space] and you'll see 'Paused' appear in the upper-right corner.

Next, select menu option Time > Set Time - leave the TimeZone as Universal Time and enter Julian Date as 2456853.10696 and hit Tab.  This should set the Date and Time to 14 Jul 2014 17:34:01 without all the fiddling with the entry fields.  Then click [OK]

Next, let's find our latitude and longitude position over the Earth, use Navigation > Go to Objection, make sure the object is Earth and enter the decimal latitude and longitude values:
Latitude: 40.93634
Longitude: -119.09180
Distance: 117.59 km
Then click [Go To]

You should see a very blurry view of the ground, way too close unless you have installed higher resolution Earth images.

Next use Navigation > Select Object and enter 'Moon' and click [OK]

Now type the letter 'c' (for center).

You should be seeing something like this.  We can already see the ocean just about where it appears in the video and the moon at about the same height over the horizon.

Figure 6. Finding the Moon in Celestia

If you have enabled Celestia URLs you can click on this link to match the view.

The next things we need to adjust require trial and error.  Go to Help > Controls for information about how to control the view but the key controls we need are just the following:

Left-Button + Drag will Orient the Camera.
Left/Right Arrow will Roll the View
Shift-Left-Button + Drag will change the Field of View (FOV)

So go head and practice using these controls to position the Earth in the right orientation and use the FOV adjustment to make the moon the right size to match the image from the video.

We can already see that the view comports pretty well with the video.  The Earth appears more curved in Celestia because it doesn't have atmosphere, glare, exposure, and lens distortion to worry about - but we will get to that...

Here is the image I ended up with after rotating around and tweaking my view.  I scaled both the original and Celestia to fullscreen and tried to 'roughly' align everything.  I'm not worried here about being a few pixels off.

Figure 7. Orient and set Field of View to match photo

The CSXT Video

Now I want to compare our information to the images we see in the video itself.  Of course, they use a GoPro which has a wide-angle lens and to keep things cheap they use a "fish-eye" or curvilinear lens instead of a much bulkier and expensive rectilinear lens (a good wide-angle rectilinear lens will cost you around $2000-$3000 and is fairly large and heavy due to all the corrective elements required) which means we will need to deal with that.  However... we can easily convert between rectilinear and curvilinear with software.  For our purposes here I'm going to use a free, online correction tool called Kako so that you can replicate this yourself.  Photoshop has a whole lens database and can correct all kinds of lens defects.

So here is my raw image from the video at 1:23 when the horizon is fairly level with the camera.  Of note here is that the horizon is still showing positive curvature even with the horizon is well below the center of the lens.  Flat Earthers love to complain about fish-eye lens but in the next image we will see that the fish-eye effect is hiding the true curvature here.

Figure 8: Raw Capture From CSXT Video circa 1:23

So now we can go into Kako and upload this image and correct the fish-eye distortion (upload your image, select Lens 18, and click [Config] button), giving us our roughly rectilinear image (and we can see that correcting for the GoPro Lens we have significantly more apparent curvature):

Figure 9. Frame capture from CSXT Video corrected for GoPro curvilinear lens distortion

And finally we can overlay this image with the Celestia image above and see that we have a very good match between the shape of the horizon and the position of the moon and the ocean.  My Celestia moon was a few pixels too small in the end so I probably need to slightly zoom in to get a slightly better alignment.

Figure 10. Celestia and corrected CSXT images overlayed

When you make a Flat Earth model that is this accurate I'll worry about getting it perfect.  But WOW - We took a rough latitude, longitude, time (from TWO years ago), and altitude and we very closely matched a photograph from space all using "Globe Math".  I love it when a plan comes together.

Flat Earth Model?

Now let's look at this data on the favorite 'flat map' of the Flat Earth community, the Gleason map.  If you have some other 'flat map' you would like to use please share with the rest of us.  Of course, the ugly truth of the Flat Earth movement is that they have NO working map that is both Area and Distance accurate and they have no working model that is based on a FLAT geometry which can predict sunrise/sunset times, azimuths, and elevations - much less one for lunar observations.  So this is the best I can do for them but it shows the absurdity of the Gleason map, and all similarly conceived maps quite well.

I made a white spot with red outline 'about' where the moon should be and I marked out 248.22° from the Black Rock Desert area and it ISN'T EVEN CLOSE.   It's pointing thousands of miles away from the direction of the moon.  How is this "100% Proof"?

Figure 10. Lunar position and view doesn't work on Gleason Flat Map

To me this should be an embarrassment to the Flat Earth community.  If you want to be taken seriously you should take your evaluations more seriously.

The Flat Earth community rails about how "NASA Lies" but their own community is the one repeatedly busted for faking their data, faking photographs, and even faking that 'Moon Hoaxer' film about Kuberik.

If you look at all the evidence that proves the moon landing was fake and still think its real, you are a fool. #FlatEarth pic.twitter.com/eRXiNc9xM5

— Flat Earth (@FlatEarthReal) September 24, 2016

@FlatEarthReal @FilmGmbh
When you use fake photos as your evidence, who becomes the fool? Original here. #science pic.twitter.com/8pZkZJI6qT

— Dean Hanson (@abqDean) September 24, 2016

@1stTryScience @FlatEarthReal Yup - they copied Alan Bean from this photo and pasted him on the visor pic.twitter.com/eWz9owuzRd

— محمصة الراس (@toasterhead) September 24, 2016

Additional Information

Figure 11. CSXT Team - Flickr

Figure 12. Lunar Position chart aligns with mooncalc

Ky's response on the video:

Figure 13. Ky's post on video with launch time & timezone

Flat Earth Follies: TEH Radiation!

Flat Earth Claim

We couldn't get past the Van Allen Belts because it would be too much radiation.

Analysis

Why would I do an Analysis when this awesome page already exists?

Apollo and the Van Allen Belts 
an estimate of the radiation dose received
Robert A. Braeunig

His conclusion?

Based on my analysis of electrons, protons, and bremsstrahlung, the predicted total dose received by the Apollo 11 astronauts as a consequence of their transits of the Van Allen radiation belts was only about 32 mrem, or 0.016 rads (all from protons ≥100 MeV). This shows that the Apollo trajectories though the VARB were not only survivable, but that the radiation doses received were inconsequential. Of course the VARB were not the only source of radiation to which the crews were exposed. To record the actual skin doses, the astronauts worn dosimeters. These dosimeter measurements for all the Apollo missions are summarized in Table 8 (Apollo 7 and 9 were Earth orbit missions).

Flat Earth Follies: The Incredible Shrinking Mountain

Here is a picture of the majestic Mt. Rainer which rises 14,410 feet above the nearby Pacific ocean, taken from Lake Washington, about 55 miles / 88513.9 meters away (circa 47.604669°N 122.235966°W)

Associated Press, 2006

The next image is also of Mt. Rainer but taken from about 130 miles / 209215 meters away over the Strait of Juan de Fuca by djboptics.  The exact location isn't known but I estimate somewhere around Cadboro Bay based on the view of the mountain (circa 48.454580°N 123.286889°W).

Mt. Rainer, by @djboptics

 Not only does Mt. Rainer appear smaller here, due to the effects of perspective (and camera field of view), but by scaling the images and matching them up we can see that a substantial fraction of the mountain is missing!

Mashup done by @NbZh_

It is very plain to see the effect of Earth curvature here on the apparent height of the mountain.

You don't need any math or training to see that half the mountain has gone missing and it's not from the effects of perspective or else the top of the mountain would be smaller in proportion as well.

But let's do a little math anyway and see if we can make sense of these observations!

To the Math!

In order to size up these two views of the mountain I'll need to know a few things.

First, what does MSL mean when we say the mountain is 14,410 feet "Above Mean Sea Level"?  And what is our basis for MSL at each location. Mean Sea Level is simply the elevation which we believe a static ocean would form by carefully removing the effects of tides, winds, and other external factors.  This is the elevation which we agree to call zero, and since common elevations are given in relation to this value we need to understand what this zero value means before we use it improperly.

The basis we're going to start with here is the Earth Ellipsoid model as defined by WGS84.  The WGS 84 datum surface is an oblate spheroid (ellipsoid) with major equatorial radius a = 6378137 m and flattening f = 1/298.257223563.  From these two values we can get the ellipsoidal radius of curvature for each of our latitude points, this is where our zero value lies.  When we are looking over larger distances we have to take into account the fact that our zero point is NOT uniform over that distance.  Indeed, we will find there is a significant change in the zero elevation over this distance, so this ellipsoidal shape must be taken into account to get accurate measurements of distant objects.

We can use the equation below to find our approx radius of curvature given our latitude θ.

a(1-f+sin²(θ))

Mt. Rainer is located at  46.852305°N which gives a radius of 6366753.797 m.
Lake Washington view was located around 47.604669°N which gives us a radius of 6366473.847 m.
And our last observation was around 48.454580°N which gives us a radius of 6366158.416 m.

So the radius at Lake Washington was 279.95 meters lower than at Mt. Rainer itself, this is because Mt. Rainer lies South of our location and the Earth is getting wider and wider as we move towards the Equator.  And the radius at our Northern-most point was 595.38 meters lower.

So we will need to take these into account when we estimate our view of the mountain.

The next equation that is needed is to find the angular size of a distant object.  This is pretty straight-forward but might seem a little confusing at first - this is based on finding the angle of a right-triangle since we know (roughly) the distance and the height of the distant object.  Our equation is:

AngularSize = 2 * arctan( ObjectSize/2 / Distance )

So we take one-half the size of the object and divide that by the distance, and then double that one-half angle result to get our full angle.    This inner part ObjectSize/2 / Distance just gives us a Slope and then arctan() function converts that into an angle.  That's all that is going on.  For example, a slope of 1 equals 45°.

Wait, we're on a Globe and the distance is the Great Circle Distance

One possible challenge here is that our ground distance is not over a flat surface.  Since we're only talking about 130 miles here let's take a quick estimate based on a circle using the half-sin rule, where R is the radius at our midpoint, arc length s is our 130 miles and we want to find a, chord length:

s = R*θ
a = R*2*sin(θ/2)

Solve the first equation for θ = s/R and substitute that in:

a = R*2*sin(s/R/2)

That gives us:

a = 6366473.847*2*sin(209215/6366473.847/2) = 209205.586 meters = 129.994 miles

Well, that was pointless.  At least we know that 130 miles is a VERY TINY chord on the total curvature of the Earth and our estimated location is probably wrong by more than .006 miles so we can say that our measurement error here is greater than the actual difference so we can ignore this.  However, as you move further away this would increasingly become more significant.

How tall should 14,410' / 4392.168m appear to us?

Ok, so we can now estimate the total angular size of the mountain, let's work in meters, and we will add in our Ellipsoidal delta (because that raises the mountain up a bit) and subtract out the apparent elevation shift expected from Earth's curvature (because that would be lowering the mountain over our horizon).  Curvature is approximately 615.37 m and 3438.53 m respectively for these distances.

2 * arctan ( (RainerHeight+EllipsoidalDelta-Curvature)/2 / Distance)

We do this twice, since we have views from two different distances.

Lake Washington:

2 * arctan ( (4392.168+279.95-615.37)/2 / 88513.9) = 1.2°

Cadboro Bay (approx):

2 * arctan ( (4392.168+595.38-3438.53)/2 / 209215) = 0.42°

These are pretty small angles so both images appear to have been taken a long focal length (zoomed in) which gives you a fairly narrow field of view.  For comparison, an 800mm lens in 35mm format gives about 1.7° Field of View in the vertical so we seem to be in the right ballpark.

The ratio between these two is ~2.82

Looking at our two pictures we find the height of the mountain in pixels in each one.  This is difficult to do because we can't see the horizon so I've just made a rough swag (and compared to some other photos with wider views), but we're just doing a rough estimate here (there is only so much we can tell without knowing all the exact details of the observers height, the properties of the lens, and other variables anyway).  But this gives us:

Lake Washington pixels: ~425

Cadboro Bay pixels (scaled to match Lake Washington Image): ~148

452/179 gives us a ratio of ~2.87

That's pretty good agreement with observation.  At ~418 pixels it would be the same ratio.  I do think maybe the horizon could be higher up than I marked it in the below image:

Anyway, thanks to NbZh_ for the mashup and djboptics for the original image.  I just wanted to take it a step further and see if these views aligned well with the Earth model and they seem to align pretty well.

Flat Earth Follies: The ISS is going too fast to take unblurred pictures!

Flat Earth Claim:

The ISS is going too fast to take unblurred pictures!

The Facts:

This is what comes from people being mathematically illiterate.

The orbital velocity of the ISS is about 17,150 miles per hour or about 4.76 miles per second.  That sure SEEMS fast, golly gee.  But if that is where you stopped in your thought process then you will come to false conclusions because you have ignored critical variables in the question.

The main variables in photographic exposures are lighting, distance, film speed, shutter speed, and aperture.

From this article in PetaPixel we find that they commonly use the 'Sunny 16' rule which says:

On a sunny day set aperture to f/16 and shutter speed to the reciprocal of the ISO film speed/setting for a subject in direct sunlight.

So at f/16, ISO 200 the resulting shutter speed is stated as "often 1/200".

So we take our 4.76 miles per second and divide it by 200 which means the ISS moves approximately 0.0238th of a mile over that 1/200th of a second or about 126 feet.

Let's double check that:  (17150 miles * 5280 feet/mile) per hour / 60 / 60 / 200 - 125.766 feet

[126 feet already doesn't seem nearly as much of a problem]

Ok, so we've moved 126 feet but we're 250*5280 = 1,320,000 feet from the CLOSEST point to our subject but we are 7,529,000 feet from the horizon (almost 6 times further away).

So the question is, assuming we are looking directly towards the Earth, how much does that move in our visual field during that 1/200th of a second.

So the only thing remaining here is understanding how angular size works.  The good news is that it is simply based on a the consideration of a right triangle:

We're going to consider one-half of our motion on one side, and we will double that to get the full result.

So let's call the full 126 feet 'g':

g = 126 feet

So one-half that distance, or from point B to point C making side a = 126/2=63 feet

a = g/2 = 63 feet

and side b is 1,320,000 feet.  What angle α results?

We know from basic trig that

tan(α) = a/b

What this means is that a/b is our slope (simply rise over run).   What the tangent function (tan()) does is convert an angle into a slope and the opposite function is called arctan() which converts a slope back into an angle.  So we can rewrite our equation, by applying arctan() to both sides giving us:

α = arctan(a/b)

Then we substitute 'g' back into it:

α = arctan(g/2/b)

And remember that this is only one HALF the total so we double this angle to get our final angular size equation which relates this angle to the size of the object and the distance.

α = 2*arctan(g/2/b)

Plugging in our values we get:

α = 2*arctan(126/2/1320000)

So we find that this is:

0.005469° or 19.69" (less than 20 arcseconds).  This is a microscopic angle.

Since our total Field of View is on the order of 60 degrees (or 216000 arcseconds) over some 1920 pixels you have some 112.5 arcseconds per pixel.

And while our FOV isn't divided up equally over all the pixels this is good enough to tell us that we are nowhere NEAR blurring, even without getting into the detailed specifics of the lens.

Conclusion

So it seems that the amount of blurring here would almost imperceptible, even at 4K resolution.

And if they were getting any blurring f/16 is a very small aperture, there is plenty of latitude to open up that aperture a bit and increase the shutter speed, to 1/250th, 1/500th, or even 1/1000th of a second.

As you get further and further away from directly below them the distances become much larger and the angular motion less and less.  So, for most of the Earth, you are looking at only a few arcseconds over the exposure time.

Flat Earth Follies: Planes would have to constantly pitch down to fly!

Flat Earth Claim

Planes would have to constantly pitch down to fly!

Analysis

This usually starts out with a Flat Earther pointing out the 8" miles squared formula and how many bazillions of miles the plane has to drop out of the sky to fly the curve - imagining that the pilot has to constantly shove the controls further forward in a fever pitch to avoid flying off into space with their Attitude Indicator spinning like a fan.

That image is, of course, pure nonsense - even if you remove my slight hyperbole.  And given how loudly they shout that this PROVES the Earth is flat my hyperbole might even be understated.

First, please read my blog post on what 8" times miles squared means and how it was derived.  The important thing to note here is that while it is accurate enough for distances from about 1 mile to about 100 miles it is not an accurate formula for the curvature of the Earth on smaller or larger scales. It is and always was an old APPROXIMATION used by surveyors.  It was 'good enough' for what they used it for and I'm fine with using for distances between 1 mile and 100 miles.   However, it is clearly not a linear formula so you can't say well..  8" / 5280 = drop over 1 feet NOR can you go the other direction in a linear fashion (as Jeranism tries to do, his method would have the shape of the Earth being a giant wedge with constant slope).   It should be obvious (once I point it out) that 8"*d^2 is a parabola and NOT a circle.

[better approximation formula, use same units for both, R=Earth Radius:  √[d² + R²] - R ]

But what is TRUE is that the curvature drop over about 1 mile is about 8 inches.

But what is also TRUE is that the drop over the NEXT MILE is ALSO JUST 8 inches.

WHAT? How can that be?  8"*4 = 32 inches so it should be another 24 inches!?  RIGHT?

No.  WRONG.  Look at the geometry again...  (we're looking at the triangle on the right in this case)

The formula assumes your distance is ALWAYS ALONG THE SAME TANGENT TO THE SURFACE.

However, once you move YOUR HORIZON HAS ROTATED!!!!!!

Planes don't fly out 1 mile, jump down 8 inches, and fly some more - that's silly.  Planes constantly FLY THE CURVATURE because that is the ONLY line of equipotential gravity!

Over that 1 mile they would CONSTANTLY AND CONTINUOUSLY rotate at a rate that is too slight to measure and, in fact, the constant buffeting by variations in the atmosphere BURIES THAT SLIGHT ROTATION INTO NOISE.  Any variation from this equipotential would (and does) result in the plane climbing or descending, which would immediately show up as a change in the energy configuration of the plane.

Ok, that is the basics of why this argument is just wrong.  Now I'll add 1 level of detail on the major points.

What does up and down even MEAN?

I know Flat Earthers deny that Gravity exists but when we drop an item it falls to the ground - that IS GRAVITY.  It really doesn't matter WHY you think it happens or what you call it.  For our purposes here it doesn't matter -- all that matters is that you agree if you dangle something so that it can pivot freely that the little thing hanging down will point DOWN towards the ground.  Can we agree on that?  If I tie a rock to a string and hold the string the ROCK will be closer to the ground.  Right?

We can do this in a vacuum so we KNOW it has nothing to do with air pressure.

And we also KNOW that it has nothing to do with buoyancy, because buoyancy is powered by gravity in exactly the same way our rock is, without the force of gravity there is no density gradient.  We know this because we can observe the behavior of density in the absence of this downward force we call Gravity.  Here is what density does when you remove the force of gravity, things no longer 'sort' by density:

You can also spin things really fast and find that the density gradient follows the resulting vector of combining the spin vector (centrifugal acceleration) with the gravity vector.

And no, YOU WILL NOT FLY OFF THE EARTH DUE TO EARTH SPIN RATE.

How much rotation?

Ok, so how much rotation are we talking about here?  Surely the pilot would notice, right?

The Earth is ~3959 miles radius and there are 360 degrees in a full circle so that means we have approximately 360/(2*3959*pi) = 0.01447° per mile.

Let's say we're in a 747 going 500 mph, so the plane needs to rotate 500 * 0.01447° per hour, or 1/3600 of that per second

(360/(2*3959*pi))*500/60/60 = 0.00201°

So that's a constant VERY SLOW PITCH rate of 0.00201° per second or just 0.12° per minute.

But does the pilot actually has to keep pitching forward?

No.  Again, you have to understand how aircraft work.

Planes control pitch using a control surface called the Elevator (usually on the trailing edge of the Horizontal Stabilizer)

And along the Elevator you'll find another smaller control surface called the Trim Tab.

Here is the most important point in this discussion:

The Elevator does not directly control the Pitch of the Airplane - it controls the RATE OF PITCH.

I think that Flat Earthers don't understand this concept.

To maintain level flight the pilot must find BOTH the elevator trim and power setting which maintains a constant altitude - they mostly use the Vertical Speed indicator to make fine adjustments to elevator trim to find the constant pitch rate that keeps Vertical Speed near zero and then makes POWER SETTING adjustments as needed to hold that Vertical Speed with a fairly constant airspeed.   If you want a higher airspeed you need to both increase power AND adjust the elevator trim so the pitch rate matches OR ELSE YOU WILL START CLIMBING.

This is a fact, I have personally flown small planes and they teach you about power control pretty much from Day #1.  You climb & descent mostly by changing the power setting (which changes when you make other configuration changes such as increasing flaps).

Now, how is a pilot supposed to tell that a TINY fraction of the elevator trim has to do with the curvature rate as opposed to all of the other forces acting on the airplane?  They couldn't possibly.

A curved gravity equipotential presents ABSOLUTELY ZERO issues for an airplane in flight.

So YES, the PLANE is (technically) constantly pitching forward as it flies the curvature of the Earth.  But it doesn't feel like pitch because DOWN is changing at the same time and it's an incredibly slight rate of pitch overwhelmed by other dynamic forces acting on the aircraft.

What about this FE Meme about the SR-71?

Wow - there are so many things absolutely wrong in this meme I almost don't know where to start.

Where did they get 2193.13 MPH from?  I get 668.9 mph for the speed of sound at 85,000' and SR-71 nominally goes Mach 3.2 (it can go slightly faster for short periods), that gives me 668.9*3.2 = ~2140.48 mph at 85,000'.   I guess that is "close enough", so I'll use their 2194 mph for this example.  But we're also at 85,000' so I'll add that into our radius of curvature (makes very little difference).

And where did they get '1/4 MILE OF CURVATURE EVERY HOUR' from?  Wow, that is just stupidly wrong.  At 2194 mph, you go 2194 miles in 1 hour and that gives 696.83 Miles OF CURVATURE DROP not 1/4 mile!  *I've assumed no wind so 2194 mph ~ ground distance and I've used CURVED GROUND DISTANCE for the curvature calculation: r/cos(d/r)-r

And even if it was a 1/4 mile 5280/4/60 = 22 FEET, not 23.  Maybe they rounded the .0?  They clearly cannot manage division.

But their worst error is assuming that 1/4 mile of curvature MEANS 22 feet per minute of vertical speed in the first place.  As I've already shown VERTICAL SPEED WOULD BE ZERO AT EVERY POINT.

The curvature 'drop' is BUILT INTO the flight dynamics - the plane isn't flying a linear path and having to drop down, it is flying along the curve -- you have to compute the integral.

The pitch RATE would be:

(360/(2*pi*(3959+(85000/5280))))*(2194)/60 = ~0.527° per minute (~0.00878°/sec)

Still only a very slight rotation that wouldn't be specifically noticeable and would very easily be compensated for in pitch rate - in this case, by the computer that is actually flying the airplane.  But again, the computer doesn't even have to understand the Earth curvature to do this, it just needs to keep VS near zero which is determined by pitch rate and power settings.

YOU DO NOT HAVE TO KEEP ADJUSTING FOR THE CURVATURE - THE RATE OF CURVATURE IS NEAR CONSTANT.

So it isn't the 696.83 miles of curvature drop you need to worry about, let's look at that over 1/1000th of a second:

The plane moves forward ~3.218 feet
The plane pitches forward 0.00000878°
The plane DOES NOT CLIMB 0.00000024669 of a foot (the 'drop' over that distance)

After 1 second the plane HAS NOT CLIMBED 0.24667 feet but it is entirely INACCURATE to say that it has DROPPED 0.24667 feet.   If you take 0.00000024669 * 1000 you'll note that you get 0.00024669  instead of 0.24667 feet -- you are a OFF BY ONE THOUSAND TIMES HERE by assuming this is a linear function.  Every microscopic bit of pitch rotation in between ALREADY ALTERED OUR COURSE.

YOU CANNOT TREAT THESE AS LINEAR AMOUNTS THAT YOU CAN SIMPLY MULTIPLY AND DIVIDE.

It is the sum total of these CONTINUOUS microscopic adjustments that result in the curved flight path.

That is what these guys don't understand.  With the atrocious math they cited in this Meme is that any surprise?

What about the Attitude Indicator?

Gyroscopically driven Attitude Indicators are TIED TO GRAVITY.  They constantly adjust themselves to find the vertical axis using gravity to force gyroscopic precession into the vertical.   They adjust very slowly and evenly and constantly but they do adjust and if you hold the plane in a non-level attitude they will even give you the incorrect attitude as a result.

Here is a very detailed video showing how the Artificial Horizon works.  At 4:00 you can see the 'pendulous vanes' that open and close little air vents which cause the gyroscopic precession until the vanes close the vents by hanging straight down under the gyroscope.

The curvature rate is well below the rate at which the gyroscopes precess to remain in the correct orientation.

The Horizon Rises to Eye-Level

No, it just does NOT do any such thing.  Aiming a camera to put the horizon in the middle or cropping an image to put the horizon in the middle does not demonstrate any such thing - in fact they are obvious lies by Flat Earthers.

Here is a non-professional instrument (Theodolite app on phone) clearly showing the horizon dip:

And here is an actual aircraft cockpit showing the horizon dip in LEVEL FIGHT.

Professional pilot Wolfie6020 also just posted a video on this same topic (and covers a few points I didn't and has some great examples):

Conclusion

I think that about wraps it up for this Flat Earth Folly.

Planes are not 'dropping' 8 inches every mile (per se), they are flying along the constant curvature of the gravity equipotential, while constantly adjusting pitch ever-so-slightly by means of the elevator trim setting which controls the pitch RATE of the airplane.  ANY deviation from that rotation results in the plane climbing or descending which immediately shows up in the Vertical Speed indication and power settings would be adjusted accordingly.  This constantly rotates their 'tangent' so there is no 8" to drop at the end of each mile, it's a constant, smooth, and VERY SLIGHT curve that presents no problems for pilots, and would be virtually undetectable in the face of other forces acting on the airplane, even at 500 mph.  The plane is simply tweaked for near zero vertical speed and that's all that is required.

The Attitude Indicator is very clearly compensating for all kinds of precessional forces acting up on it and to remain accurate over the longer term must be tied to the gravity potential as well (so it remains vertically aligned over time, reacting only to sharper movements of the airplane over the short term).

Flat Earth Follies: Blowing Gravity Out Of The Water!

Well, that is what this guy claims...



It's not gravity you see -- it's because fluids magically seek their own level.

There is just one little problem... When you remove gravity the fluids no longer know which way is down and no longer have the magic to 'seek their level' just due to air pressure.



So he's just Flat Out Wrong.

Flat Earth Follies: LAX to LHR Flight Path Takes You Near Iceland

YouTuber 'Flat Earth Addict' (a good term for them, they know it's bad for them but they can't quit) wonders why a flight going from the UK to LA passes near Iceland?

NO DUH

BECAUSE THAT IS CLOSE TO THE SHORTEST PATH ON THE GLOBE

Because both points are well into the Northen Hemisphere his line on Gleason's azimuthal equidistant projection (projection FROM  A GLOBE, LOL).  Beyond that point the Gleason map IS TERRIBLE -- sizes, distances, compass headings, everything is far worse than on the Mercator projection.  The ONLY thing accurate on the azimuthal equidistant projection is distances TO THE CENTER POINT.  All other distances are WRONG and they get more and more WRONG as you go further 'South' (towards his outer edge).

Notice how the paths are ALMOST identical?  What is this guy trying to prove exactly?

You can find the story, such as it is, here:

Why did American Airlines flight travel 1,000 MILES back to London instead of landing in Iceland after medical emergency?

and a little more detailed here.  (sounds like maybe aerotoxic syndrome)

Ok -- so we don't really know much of anything.  Maybe the pilot didn't think the medical issue serious enough to strand everyone else in Iceland so back to LHR seemed the best choice?  So what?

All this proves is that they likely were close to Iceland which fits perfectly well with the Spherical Model Earth.

FlightAware shows the actual flight path on the Mercator projection with a dotted line taking it a little closer to Iceland than a straight-line distance -- this is not usual for many reasons including safety, flight congestion, and legal.  The line isn't straight on the Mercator projection because it PURPOSEFULLY distorts areas away from the Equator - BUT it does this to preserve compass directions.