1

Flat Earth Theory / Vendée Globe

« on: November 11, 2020, 10:36:24 AM »

Two days ago, 33 sailors have set sail from Les Sables d'Olonne, France. The Vendée Globe is a solo non-stop round-the-world race, along the Clipper route.



It's a tough race: typically, only half the sailors make it to the finish line.

https://www.forbes.com/sites/tmullen/2020/11/09/the-insanity-and-elegance-of-the-vende-globe-sailing-race

https://www.espn.com/olympics/story/_/id/30249677/bubbles-separation-solitude-trying-cope-2020-let-examine-vendee-globe

Their progress can be tracked online : https://www.vendeeglobe.org/en/tracking-map

On a flat bipolar Earth, this kind of circumnavigation is not possible. On a flat monopole Earth, it could be possible, but the southern part along the roaring forties would be much longer.

Are these sailors and the race organization part of a conspiracy?

2

Flat Earth Theory / Angular size of the Moon

« on: October 12, 2020, 11:06:29 AM »

The angular size (apparent diameter) of the Moon varies with time. Its variation is somewhat faster than the phases cycles: sometimes the largest apparent Moon will coincide with a full moon, causing the so-called "Supermoon"



The difference between a small (apogee) Moon and a big (perigee) Moon is about 13%. Examples are easily found:







Pictures typically depict a full moon because that's more spectacular and makes it easier to compare, but they could be made with any visible phase.

This kind of photos is quite easy to make, photographing the Sun is tricky and even dangerous if done wrong, but for the Moon you only need a decent DSLR and long-focus lens to get this kind of pictures.

Two things to note:
* When the Moon appears bigger, it appears bigger for everyone, wherever they are on Earth
* The angular size of the Moon doesn't vary significantly as it moves through the sky

Now, from the wiki (https://wiki.tfes.org/Moon) :

The Moon is a revolving sphere. It has a diameter of 32 miles and is located approximately 3000 miles above the surface of the earth.

Using these values, for a Moon that's directly overhead, we calculate an angular size of about 0.611 degress or 36.6 arcminutes (https://rechneronline.de/sehwinkel/angular-diameter.php ). That's slightly above the observed largest Moon but in the right ballpark.

But for a Moon that's above a point 5000 miles away, the distance from the observer is sqrt(3000^2+5000^2) (Pythagorus) = 5830 miles. That gives us an angular diameter of 0.314 degrees or 18.8 arcminutes. Half as small. Such a variation has never been observed anywhere, not even at different times, and certainly not from two places at the same time

On a flat Earth with a close Moon:
* What causes the variation of the apparent size of the Moon for every observer on Earth over a cycle of about 27 days, slightly shorter than the cycles of lunar phases?
* What causes the Moon to have an apparent size that's not significantly different whether it's 3000 or 6000 miles away (without ever distorting its shape)?

3

Flat Earth Theory / Is there a flat Earth Jean Meeus?

« on: September 20, 2020, 09:44:16 AM »

Jean Meeus is a Belgian astronomer. He published a book called Astronomical Formulae for Calculators and another called Astronomical Algorithms. As their name suggest, they detail ways to implement formulas and algorithms to calculate position of celestial bodies and other astronomy-related calculations.

His algorithms have been implemented in a variety of languages:

https://developer.aliyun.com/mirror/npm/package/meeusjs
https://www.npmjs.com/search?q=keywords:meeus
https://crates.io/crates/meealgi
https://pypi.org/project/PyMeeus/
http://libnova.sourceforge.net/

His formulas and algorithms use angular values: latitude and longitude for terrestrial location, declination and right ascension for positions of celestial bodies. They use values such as the eccentricity of the Earth's orbit or the obliquity of the ecliptic, that only make sense in a spherical Earth model. They are, most definitely, built on a round Earth model.

There is a live implementation here of the formulas for sunrise and sunset times: https://www.esrl.noaa.gov/gmd/grad/solcalc/index.html and details here https://www.esrl.noaa.gov/gmd/grad/solcalc/calcdetails.html , you can even download a spreadsheet to run it on your computer with Excel or OpenOffice.

Anyone can check that the calculations are pretty accurate. The sunrise and sunset calculations are just a small subset of Meeus' algorithms and formulas, but they are an obvious example because everyone will know what they mean and will be able to check their accuracy.

Is there a flat Earth equivalent, that works equally well, based on EA or anything else?

4

Flat Earth Theory / Moons of Jupiter and the speed of light

« on: May 03, 2020, 12:07:19 PM »

Jupiter has many natural satellites, but only four of them are so big and bright they were identified by Galileo more than four centuries ago. They are now known as the Galilean moons: Io, Europa, Ganymede and Callisto.

Even with a small telescope, it's possible to witness from Earth the Jovian equivalent of our eclipses: moons casting their shadows on Jupiter, moons being eclipsed by the shadow of Jupiter. See https://www.space.com/6705-moon-shadows-jupiter.html :

"The shadow of the moon, cast by the sun, can also be seen to transit across the cloud tops. Because the shadows are intensely black, they are much easier to see than the moons themselves.

When a moon passes behind the planet, it can be occulted by the planet itself, or eclipsed by Jupiter's immense shadow. You can even watch a moon emerge from behind Jupiter, glow brightly for a few minutes, and then fade from sight as it enters Jupiter's shadow."

A number of sites give the times for each event (occultation, transit, shadow, eclipse), for example:
http://shallowsky.com/jupiter/
https://skyandtelescope.org/wp-content/plugins/observing-tools/jupiter_moons/jupiter.html

Jupiter is so big and bright it can even be seen in daylight conditions: https://www.skyatnightmagazine.com/advice/skills/jupiter-moons-double-transit/

Depending on where Jupiter, the Earth and the Sun are, eclipses will be seen after or before occultation. A few months after opposition, the two outermost Galilean moons (Ganymede and Callisto) can be seen to disappear behind Jupiter, reappear, and then disappear again in Jupiter's shadow: https://books.google.com/books?id=w9URBwAAQBAJ&pg=PT133&lpg=PT133&redir_esc=y#v=onepage&q&f=false . The same phenomenon cannot be observed for Io and Europa because their orbit is too close to the planet: the eclipse starts before the occultation ends or the occultation starts before the eclipse ends.

Variations in timing of these eclipses were used by Danish astronomer Ole Rømer to make the first estimation of the speed of light in 1676, when at that time light was thought to propagate instantly and have an infinite speed. Rømer's first estimation was too low, but definitely in the right ballpark (see also http://www.phy6.org/stargaze/Sun4Adop1.htm ). The same method was later used to make more refined estimations that were in line with the current accepted value of c.

Now, on a flat Earth:

1) Is our Sun the source of light for the Jovian system?
2) If not, then what is it?
3) In any case, how far is the light source from Jupiter and its satellites?

5

Flat Earth Theory / Comet Shoemaker–Levy 9

« on: April 25, 2020, 10:48:30 PM »

In 1993, a comet was discovered. Calculations showed that it would collide with Jupiter in July 1994.

As the fragments of the comet collided with the planet between July 16 and July 22, even amateur astronomers with small telescopes could see the effects of the impact, which were at least as visible as the famous "red spot".

How could such a prediction be made, more than a year before the event, without very precise knowledge of the orbits of Jupiter and the comet? How does that fit with a flat Earth view claiming that little is known about the celestial bodies and their distances, and predictions are only based on patterns? This collision was the first one ever observed between two bodies of the Solar System, it's hard to find a pattern when there's a single occurrence.

6

Flat Earth Theory / Are flights from and to French Polynesia a hoax?

« on: April 12, 2020, 01:54:08 PM »

According to Wikipedia, Tahiti is the largest island of the Windward group of the Society Islands in French Polynesia, located in the central part of the Pacific Ocean. On the island, there is the only international airport in the region, Faa'a International Airport.

There are more, but let's focus on two flights :

Air Tahiti Nui flight TN 102: Auckland - Los Angeles via Tahiti (this route is the usual way to go from Tahiti to mainland France: as there are no direct flights to Paris, people fly to LAX and from there to CDG)

LATAM Airlines Group flight LA 836: Tahiti - Santiago via Easter Island (weekly flight, Easter Island is a popular destination for Tahitians or tourists visiting Polynesia - except of course when there is a lockdown in place)

Let's calculate the distances for each segment, and match with the typical scheduled time:
http://www.gcmap.com/mapui?P=AKL-PPT-LAX
http://www.gcmap.com/mapui?P=PPT-IPC-SCL

AKL-PPT: 2,544 mi, 4:45h
PPT-LAX: 4,095 mi, 7:41h

PPT-IPC: 2,644 mi, 5:05h
IPC-SCL: 2,335 mi, 4:30h

Now, how would that work on a flat Earth?

On a bipolar model, it's simple, it doesn't work at all. You can't go from Auckland to Santiago via Papeete and Easter Island over the Pacific Ocean. At least one of these flights cannot exist.

On a "standard" monopole, such as the popular azimuthal equidistant projection, the segments could be possible but the distances don't match at all: Tahiti appears closer to Los Angeles than to Easter Island or Auckland, while the PPT-LAX flight actually takes approximately 50% longer, which is consistent with the calculated great circle distances. Moreover, if you sum Auckland-Papeete, Papeete-Easter Island, Easter Island-Santiago, it would take just over 14 hours. Compare the sum of distances on any flat Earth monopole map to the distance of a direct flight from Rome to Los Angeles that takes 13 hours. Even without stopovers, Auckland-Santiago is more than twice as long. These flights would need to be supersonic.

Conclusion, either the information on these flights is false (they don't exist at all, or their duration is seriously altered), either none of the proposed flat Earth maps is correct.