Debunked: "Celestial navigation is based on elevation angles from a Flat Earth"

[Mendel]

You seem to have come up with a number for "very very far", as that's one of the inputs necessary for getting a number out.

No. On the globe, the main change is tilt, which depends on the observer's position, and time, not the distance to the stars.

As your parenthetical "infinitely" indicates, you're justified in just taking the limit as the number you chose tends to infinity, and that leads to a tilt of 0⁰ in the limit (which is well behaved). You were right at the outset, the adding of numbers has fuzzied things.

In empirical practice, the width of Earth's orbit needed to be utilized to discover stellar parallax with astronomical instruments. That means, apart from tilt, the sky is pretty much the same (for household purposes) everywhere on Earth.

"All finite things are but nothing to the infinite" - Epifatphilus, famous non-existent non-greek non-philosopher.

"The finite is annihilated in the presence of the infinite, and becomes a pure nothing. " — Blaise Pascal, On the Wager for God's Existence

 

[FatPhil]

No. On the globe, the main change is tilt, which depends on the observer's position, and time, not the distance to the stars.

There was no globe!? There was no tilt!? Did I misread something?

In empirical practice, the width of Earth's orbit ...

OK. I misread something. There were no orbits in what I read.

"The finite is annihilated in the presence of the infinite, and becomes a pure nothing. " — Blaise Pascal, On the Wager for God's Existence

Damn! I'm gonna have to re-evaluate how smart that Fatphilistarchus was.

 

[MyMatesBrainwashed]

My brain got a bit tired of infinity.

And I realise that none of what I was discussing does much to the "navigation angles based on a flat earth" point.

Then I got onto thinking about a flat earth with a horizon that rises to eye level and what that means about elevation angles measured from the horizon.

But my brain coped with that worse than it did infinity and I had to give that up too.

Maybe someone else can theorise on it better.

 

[Scaramanga]

There are various ways to check the validity of this claim.

Noooo. I'm not going to waste even one attosecond of my life checking the validity of any flat Earth claim. In many ways...I think people keeping on 'explaining' things to flat earthers is precisely what gives them a sense that they have something worth explaining. They don't. People believe in flat Earth because they are stupid. I mean...let's finally say that and move on.

 

[Mendel]

I missed this the first time around.

The first premise is commonly accepted by flat earthers, and the group I linked to will happily use it in their imagery, while celebrating their knowledge of this formula. This is one of the images they use:

to anyone who's mastered trigonometry, it's clear that the "flat" version of this must involve the sinus of 54⁰. The fact that it doesn't proves that the 3240 nautical miles is in fact an arc length. Arc lengths are directly proportional to angles, straight ("flat") distances are not.

I learned a lot of interesting stuff while explaining to Flat Earthers—though rarely from them.

 

[Breadloaf]

Some youtubers claim that celestial navigation is based on elevation angles from a Flat Earth. For example:

https://www.youtube.com/c/NathanOakley1980c/videos

Not only is celestial navigation allegedly based on a Flat Earth, it works so well that it amounts to proof that the earth is flat!

There are various ways to check the validity of this claim. One option is to read a book about the subject, and see what kind for formulas are used to do the calculations. A nice resource that I like to use is

www.TheNauticalAlmanac.com

The website hosts many PDF files with formulas and explanations.

Understanding the maths behind all of this is challenging. My idea for this debunk is to use data from multiple screenshots from my Nautical Almanac App (link: https://play.google.com/store/apps/details?id=com.skrypkin.nauticalalmanac) to show that the data from my app contradicts the claim. Using a combination of times and dates and positions that I found, only simple arithmetic operations are needed to do a sanity check on the output of the program.

I will use 2 main premises to analyze the numbers from the app:

• The distance to a Geographical Position (GP) is the Zenith Distance multiplied by 60 nm. The zenith distance will be calculated as 90° – Height. Note that 'Height' means 'Elevation angle' in this context.
• If the difference in Azimuths is 180° then the distance between 2 GP's can be calculated by adding the two distances. This is trivial, and follows from the Law of Cosines.

The first premise is commonly accepted by flat earthers, and the group I linked to will happily use it in their imagery, while celebrating their knowledge of this formula. This is one of the images they use:

View attachment 51039

And this is a modified version of the image, which illustrates the type of calculation I will use:

View attachment 51040

I took 12 screenshots from my Nautical Almanac app. They involve 4 Geographical Positions (GP's):

View attachment 51057

My app won't show the Latitude of Longitude of the substellar points, but it will show the Greenwich Hour Angle (GHA) and Declination, from which the GP can be derived. The relationship between them two should be obvious from the table. For point D, because the GHA is greater than 180°, the longitude East is calculated by 360° - 234° 45.92' = 125° 14.08'.

Each of the 4 GP's is associated with a moment in time when a star is directly above that position:

View attachment 51058

In celestial navigation, it is common to calculate the Height (=elevation angle) and Azimuth of a star from a certain Assumed Position and time. I have chosen positions halfway in between each combination of GP's:

View attachment 51060

Using Assumed Position 1 and the times for A (left) and B (right):

View attachment 51041 View attachment 51042

The difference in azimuths is 180° - 0° = 180°.
The distance to both A and B is (90° - 35° 18.78') x 60 = 3281.22 nm
This gives AB = 3281.22 + 3281.22 = 6562.44 nm


Using Assumed Position 2 and the times for A (left) and C (right):

View attachment 51043 View attachment 51045

The difference in azimuths is 222° 15.06' – 42° 15.06' = 180°
The distance to both A and C is (90° - 35° 14.08') x 60 = 3285.92 nm
This gives AC = 3285.92 + 3285.92 = 6571.84 nm


Using Assumed Position 3 and the times for A (left) and D (right):

View attachment 51046 View attachment 51047

The difference in azimuths is 317° 44.90' – 137° 44.90' = 180°
The distance to both A and D is (90° - 35° 14.13') x 60 = 3285.87 nm
This gives AD = 3285.87 + 3285.87 = 6571.74 nm


Using Assumed Position 4 and the times for B (left) and C (right):

View attachment 51048 View attachment 51049

The difference in azimuths is 312° 7.53' – 132° 7.53' = 180°
The distance to both B and C is (90° - 35° 16.41') x 60 = 3283.59 nm
This gives BC = 3283.59 + 3283.59 = 6567.18 nm


Using Assumed Position 5 and the times for B (left) and D (right):

View attachment 51050 View attachment 51051

The difference in azimuths is 227° 52.50' – 47° 52.50' = 180°
The distance to B is (90° - 35° 16.45') x 60 = 3283.55 nm
The distance to D is (90° - 35° 16.46') x 60 = 3283.54 nm
This gives BD = 3283.55 + 3283.54 = 6567.09 nm


Using Assumed Position 6 and the times for C and D:

View attachment 51052 View attachment 51053

The difference in azimuths is 270° - 90° = 180°
The distance to C is (90° - 35° 15.33') x 60 = 3284.67 nm
The distance to D is (90° - 35° 15.32') x 60 = 3284.68 nm
This gives CD = 3284.67 + 3284.68 = 6569.35 nm

In summary, we found the following distances:

AB = 6562
AC = 6572
AD = 6572
BC = 6567
BD = 6567
CD = 6569

If we try to make a map of the flat earth that this is supposedly based on, with accurate distances, once we have placed A, B and C on it, we can't find a place for point D such that the distances to the other 3 points are correct:
View attachment 51054View attachment 51056View attachment 51055



The 6 distances between these 4 points, being impossible on a flat surface, contradict the claim that a Flat Earth is being used as the basis for celestial navigation.

Because the almanac-based positions are being used in combination with spherical calculations, trying to lay them down flat without undoing that spherical logic will obviously cause distortion. I feel like this only shows that spherical assumptions don't flatten cleanly. Which was already known.

 

[Mendel]

Because the almanac-based positions are being used in combination with spherical calculations, trying to lay them down flat without undoing that spherical logic will obviously cause distortion. I feel like this only shows that spherical assumptions don't flatten cleanly. Which was already known.

The almanac data is verifiable, though. The ground point data means that at that position and at that time, that star is directly overhead. That's true no matter what you believe the world looks like.
And then you only need the calculation from the screenshot, where you multiply the (observable and measurable!) angle with 60 nm/⁰.

If this didn't work, ships in the age of sail could not have crossed oceans reliably. Even early airliners had sextant ports in the ceiling of the cockpit, for the navigator to use.

Now if you consider that this method of navigation was developed centuries before computers were even invented, it stands to reason that if Flat Earth celestial navigation was possible, someone would've explained how it works. But that's not the case. (And we can explain why not.)

 

[Breadloaf]

The almanac data is verifiable, though. The ground point data means that at that position and at that time, that star is directly overhead. That's true no matter what you believe the world looks like.
And then you only need the calculation from the screenshot, where you multiply the (observable and measurable!) angle with 60 nm/⁰.

If this didn't work, ships in the age of sail could not have crossed oceans reliably. Even early airliners had sextant ports in the ceiling of the cockpit, for the navigator to use.

Now if you consider that this method of navigation was developed centuries before computers were even invented, it stands to reason that if Flat Earth celestial navigation was possible, someone would've explained how it works. But that's not the case. (And we can explain why not.)

I'm not denying a spherical recalculation can be used to verify your position and be used for navigation. Almanac data is ultimately based on observational data of altitude angles to celestials (with celestial sphere construction based on apparent positions) in reference to a 2 dimensional plane.

Since these measured angles are taken with respect to the horizon, every baseline of altitude angle is considered to be parallel with this plane of reference. So the empirical measurements are and must be in reference to a flat plane. In this way elevation angles live up to there definition, since they are measuring a flat distance to the ground position, giving circles of equal altitude and providing a reference for (flat) distances, which in turn is a prerequisite for globular transformations.

This data is sub-sequentially recalculated and projected onto a sphere. If you then take an elevation angle in reality with a sextant and you check it on your sphere (GPS), it will work - provided that the original measurements are valid and the spherical transformation is accurately calculated and sustained within verification.

The globe model in this instance is a mathematical reinterpretation of those measurements, not their source. The model should only be considered in violation with the original empirical framework if one reifies the model into existance (in terms of topography) without undoing all the recalculations. So in this case this would be much more than a lineair scale, obviously. As you correctly stated the "maths behind all of this is challenging".

Moreover, the fact that this recalculation works for navigation verifies that the original measurement-data must be topograhically correct. Ofcourse without any assumption or claim about what the celestials are (which is irrelevant anyway, because we only use them to measure time and navigate the earth). If the earth wasn't correctly measured in the first place, you wouldn't be able to use them for navigation, whether or not you recalculate them onto a globe.

So the fact that you can still use the original data on a globe projection affirms the valididty of your original data and its framework and confirms you did a good job in terms of calculation and correction for distortions succesfully construing a spherical map based on empircal measurements that reference a flat plane.

 

[Mendel]

I'm not denying a spherical recalculation can be used to verify your position and be used for navigation. Almanac data is ultimately based on observational data of altitude angles to celestials (with celestial sphere construction based on apparent positions) in reference to a 2 dimensional plane.

No, it's not.

Since these measured angles are taken with respect to the horizon,

No, they're not. They're measured with respect to the zenith, via a plumb.

every baseline of altitude angle is considered to be parallel with this plane of reference.

Yes and no.

So the empirical measurements are and must be in reference to a flat plane.

They're in reference to the zenith.
In vector mathematics, you define a plane by giving a point (the observer) and a "normal" vector perpendicular to the plane (the direction to the zenith, or "up", as indicated by a plumb line).
This works at night, when the horizon may not be visible, or in aircraft, where it [the horizon] is not at 0⁰ elevation.

In this way elevation angles live up to there definition, since they are measuring a flat distance to the ground position,

a sextant measures an angle, not a distance

giving circles of equal altitude

no, it does not.
altitude does not come into it, and Flat Earth fails at computing the altitude of stars.

and providing a reference for (flat) distances,

no, it does not.

which in turn is a prerequisite for globular transformations.

which transformations?

This data is sub-sequentially recalculated and projected onto a sphere.

Yes, in the way I describe. The ground point of a star at 54⁰ degrees zenith angle is 54⁰×60nm/⁰=3240nm miles away.

If you then take an elevation angle in reality with a sextant and you check it on your sphere (GPS), it will work - provided that the original measurements are valid and the spherical transformation is accurately calculated and sustained within verification.

There is no "transformation" involved.

The globe model in this instance is a mathematical reinterpretation of those measurements, not their source.

Could you please explain what this means?

The model should only be considered in violation with the original empirical framework if one reifies the model into existance (in terms of topography) without undoing all the recalculations. So in this case this would be much more than a lineair scale, obviously. As you correctly stated the "maths behind all of this is challenging".

The math formulae in the almanac are challenging, but this is not.
The ground point of a star is where the star is directly overhead.
The truth of this can be verified directly, by observation, with no mathematics whatsoever.

Moreover, the fact that this recalculation works for navigation verifies that the original measurement-data must be topograhically correct. Ofcourse without any assumption or claim about what the celestials are (which is irrelevant anyway, because we only use them to measure time and navigate the earth). If the earth wasn't correctly measured in the first place, you wouldn't be able to use them for navigation, whether or not you recalculate them onto a globe.

Yes. You need a precise map, or, as ships typically carried, a globe.
Flat Earth has no map.

So the fact that you can still use the original data on a globe projection affirms the valididty of your original data and its framework and confirms you did a good job in terms of calculation and correction for distortions succesfully construing a spherical map based on empircal measurements

right!
but there's not a lot of correction for distortion because Earth is almost round.

that reference a flat plane.

wrong!
if that was true, it would be possible to do celestial navigation based on Flat Earth, but it's not.

Another simple demonstration of this is Polaris, the North star. If you move 60nm/111km south on the northern hemisphere, Polaris loses 1⁰ of elevation. This is because the direction of "up" tilts 1⁰.

If you do this on Flat Earth, the distance to make Polaris lose 1⁰ of elevation can't be constant (or the altitude of Polaris can't be constant). You cannot determine latitude with reference to Polaris on Flat Earth.

How high up is the star Polaris, near the celestial North pole? Globers say, pretty far away. FEers put it at a few 1000 km, but can't work out exactly how high. https://flatearth.ws/polaris-angle

Your "spherical transformations" claim assumes that there is Flat Earth maths that explains the observations better than ground points (observable truth!), elevation angles, and distances as ×60nm/⁰.
But the people who propose that claim can neither provide the "spherical transformations" nor the alleged Flat Earth maths that they're supposedly based on. Shouldn't that make you suspicious?

 

[MyMatesBrainwashed]

in reference to a 2 dimensional plane.

What are the 2 dimensions of the plane it references?

 

[Mendel]

What are the 2 dimensions of the plane it references?

all planes are two-dimensional, you can really choose your coordinate system any way you want
it's how planes are defined
a 3-dimensional "plane" would be a space

 

[MyMatesBrainwashed]

all planes are two-dimensional, you can really choose your coordinate system any way you want
it's how planes are defined
a 3-dimensional "plane" would be a space

Indeed.

I'm wondering where the second dimension comes into it.

 

[JMartJr]

@Mendel thanks for posting this image:

Somewhere in the files I tried to save from my most recent computer-death is a similar-but-in-the-opposite-direction chart of the Sun at the Equinox, when it is straight up to an observer at the Equator but just skimming the horizon to an observer at the poles. I though it was pretty unanswerable by itself, but when paired with this Polaris chart the degree to which the Flat Earth is palpably nonsensical ratchets up a couple of notches.

Edit to add -- don't have time to go find my version but here is a still from a Jos Leys video which is even better -- he has better graphics and also does the angles from South of the Equator... the bearings to the SUn at noon on the Equinox:

Source: https://www.youtube.com/watch?v=VHHuxMVclX4

 

[Mendel]

Somewhere in the files I tried to save from my most recent computer-death is a similar-but-in-the-opposite-direction chart of the Sun at the Equinox, when it is straight up to an observer at the Equator but just skimming the horizon to an observer at the poles. I though it was pretty unanswerable by itself, but when paired with this Polaris chart the degree to which the Flat Earth is palpably nonsensical ratchets up a couple of notches.

The really nice one is the animated GIF that bends the flat diagram into a circle, making all the rays parallel.

 

[JMartJr]

The really nice one is the animated GIF that bends the flat diagram into a circle, making all the rays parallel.

The Jos Leys videos are replete with that moment... cannot recommend them highly enough, they are beautiful as well as compelling.