The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: BoatBum on May 14, 2020, 07:27:13 PM

I have a question for the FE community. I've searched through the forums, and while I've seen posts about circumnavigating the Antarctic continent, I've not found this topic. If I've missed it  I apologize. I'm a pretty serious offshore sailor. Over 201819 I put 6,000 nautical miles on my sailboat. Since 1968, about every 4 years there are aroundtheworld sailboat races. Almost all leave from England. Some are nonstop; some have a few stops at the obvious places. The typical course is to leave England, head south to the Cape of Good Hope at the southern tip of Africa, next to Australia, then around Cape Horn at the southern tip of South America, then back to England. There's a great documentary movie, entitled "Maiden", about the first allfemale crew to accomplish this grueling feat in 1989. Using the FE model that I typically see, the distances involved in this would be hugely greater than if the Earth is really spherical. Given that racing sailboats only go about 15 mph on average, I would think that the participants would easily notice such a massive increase in the distances they have to travel. How does this reality mesh with the FE model?
I also have a question about the spherical trigonometry mathematics needed to make nautical sextants work (and they do) for ocean navigation, but I'll save that for now.
Thanks

I disagree that it would be so easy for the sailors to notice that they're traveling a greater distance than expected; there are so many xfactors involved with sailing. I suspect they rely primarily on navigation equipment to determine what kind of distances they've covered and just trust what they're told just like everyone else; for all we know innumerable sailors involved in this race were surprised that they covered as little distance as their equipment tells them, but because RE is so ingrained, they just don't question it.

I disagree that it would be so easy for the sailors to notice that they're traveling a greater distance than expected; there are so many xfactors involved with sailing. I suspect they rely primarily on navigation equipment to determine what kind of distances they've covered and just trust what they're told just like everyone else; for all we know innumerable sailors involved in this race were surprised that they covered as little distance as their equipment tells them, but because RE is so ingrained, they just don't question it.
Can you suggest a few xfactors that aggregated would account for double or triple the length required (the outside lines of longitude on the Azimuthal FE model are easily twice the circumference of the inner ones where the UK is that the race starts)? I find it very hard to believe it would be "so easy" to be mistaken about going two to three times more than expected. That would require having to go two to three times the average speed since they can literally just count the days the trip takes, right? And this would mean that lots of sailors  who are so experienced at sailing that they can travel for thousands and thousands of miles  cannot tell the difference between about 15mph and 30 or 45mph. It's trivial to tell the difference between those two speeds in a car, but then again there are things to look at on the ground. Even so, the surface of the ocean is not placid, so you can gauge it, if you've spent years on sailing boats . Plus, a great deal of the trip must be within sight of land, based on the description (certainly not all of it).
Additionally, if even a single person, out of the many who have done this since 1973 does not use sophisticated navigation equipment relying on the conspiracy, then that disproves the FE model in one fell swoop. Seems like kind of a high bar? Isn't the zetetic model all about seeing for yourself? It appears that literally anyone can do this. The website reads "Anyone, even if they have never stepped on a boat before, can join the adventure."
https://www.clipperroundtheworld.com/about/abouttherace

I disagree that it would be so easy for the sailors to notice that they're traveling a greater distance than expected; there are so many xfactors involved with sailing. I suspect they rely primarily on navigation equipment to determine what kind of distances they've covered and just trust what they're told just like everyone else; for all we know innumerable sailors involved in this race were surprised that they covered as little distance as their equipment tells them, but because RE is so ingrained, they just don't question it.
Can you not tell you're going faster when you're walking on a travelator? You know how much faster you'd need to be going to go round the flat Antarctica? It's not just a small difference.
I'm sure someone will come back to this and say there's some kind of extremely fast current that's dragging the boats round x times faster or something, well you'd only have to either stop or go the other way to find out if that's true.

I disagree that it would be so easy for the sailors to notice that they're traveling a greater distance than expected; there are so many xfactors involved with sailing. I suspect they rely primarily on navigation equipment to determine what kind of distances they've covered and just trust what they're told just like everyone else; for all we know innumerable sailors involved in this race were surprised that they covered as little distance as their equipment tells them, but because RE is so ingrained, they just don't question it.
I think if the GPS system was lying to sailors, the discrepancy between GPS and sextants and other navigation aids would be very quickly unmasked.
The earliest popular sailboat races I could find in a quick search was in 1968, ten years before GPS launches and so I have to imagine they would also have noticed when suddenly the distances changed.
There are just too many ways of measuring your position for any one to be messed with, especially in this day and age.
Even a drunk sailor would notice if a trip was many times longer than it should have been.

...for all we know innumerable sailors involved in this race were surprised that they covered as little distance as their equipment tells them, but because RE is so ingrained, they just don't question it.
A "for all we know" statement is merely a made up supposition. Take the Volvo Around the World races. If you've watched any of the footage, those guys and gals know exactly how fast they are going, where they are, where they are vectoring to, when they will get there given the variables, and where their competition is 24/7. Mishaps do occur, mostly due to the crazy routes they sometimes take chasing wind and storms (They can get up to 35 MPH, which is insane) where shoals and reefs aren't charted very well due to the remoteness of their route choice. But they still know right where they are on the planet.

I've been out sailing off shore on a sail boat too. However I was a professional merchant marine officer for the last 20 years of my career. In those 20 years at least 10 of them were spent living & working an ocean going ships transiting the oceans of the world. If you make a trip say from Shanghai, China to Long Beach, CA countless times you know within a small margin of error what the variables are. Distances are very well known. When you leave port you start the ship's engine. The ship has a propeller with a known pitch. That means for every turn of the propeller the ship travels a known distance. There's a counter on the propeller shaft and at noon every day the shaft count is recorded. You can use those figures to easily calculate a distance. Of course there's some margin for error but if you are making the same trip over & over again you know what to expect. If the distances traveled in a day don't match the known speeds it will quickly become apparent that something is wrong. You can get within a mile or two with a sextant and closer than that with GPS so any good merchant marine navigator will know when things aren't correct. I can say, in the Zetetic fashion, that the earth is proven to be spherical and can't be flat from countless ocean trips of over 5000 miles. You are wasting your time to think anything else.

I've been out sailing off shore on a sail boat too. However I was a professional merchant marine officer for the last 20 years of my career. In those 20 years at least 10 of them were spent living & working an ocean going ships transiting the oceans of the world. If you make a trip say from Shanghai, China to Long Beach, CA countless times you know within a small margin of error what the variables are. Distances are very well known. When you leave port you start the ship's engine. The ship has a propeller with a known pitch. That means for every turn of the propeller the ship travels a known distance. There's a counter on the propeller shaft and at noon every day the shaft count is recorded. You can use those figures to easily calculate a distance. Of course there's some margin for error but if you are making the same trip over & over again you know what to expect. If the distances traveled in a day don't match the known speeds it will quickly become apparent that something is wrong. You can get within a mile or two with a sextant and closer than that with GPS so any good merchant marine navigator will know when things aren't correct. I can say, in the Zetetic fashion, that the earth is proven to be spherical and can't be flat from countless ocean trips of over 5000 miles. You are wasting your time to think anything else.
Fascinating. I do have a question about the shaft count. Wouldn't that only measure speed in water (and air) that are completely still? In other words, air can push the ship a little, and water currents can also push the ship a little, right? Wouldn't this affect the measurement, or is it too insignificant?

You are correct. We also had a separate gauge that accurately measured speed thru water, but everything is relative to that water. There are some significant currents out there that can easily add 3 or 4 knots to our speed depending on our heading at the time. The current off the Japanese coast is one example. Heavy seas taken head on will also result in more turns per mile. Seas taken astern will do the opposite. You have to be familiar with the average turns per mile under calm sea conditions. Sea conditions has a bigger effect than winds. Every ship was different. You had to be familiar with a particular ship, the route, and the effects of the winds & waves on that particular ship. All that takes lots of trips to get used to what is normal. The whole idea of the shaft count in the first place was to get an idea of whether the hull was starting to get fouled. A freshly painted hull would go further with every turn than a heavily fouled one. Having said all that the GPS is the best indicator as it gives everything relative to the earth itself. Our GPS data stream drove the electronic map display and you could always see the ship's position relative to land. You could also overlay that GPS data to the radars to display the relative positions of the ship to nearby land and all the other ships either by radar return echoes or the AIS. All the deck officers have to be trained on using a sextant for navigation as well. We were required by law to carry one on board at all times in case of an emergency. We sometimes carried Merchant Marine Academy cadets on board for training purposes and they often were taking sextant sights to get up to speed using the sextant although these days it would seldom be needed. In the middle of the Pacific Ocean I could get out my iPhone and use the GPS app and obtain a GPS position that matched exactly to the main GPS receivers on the ship's console. Using all the fancy navigational equipment would allow us to accurately predict when we would arrive at the sea buoy of our destination port. This meant that we accurately knew distances & speeds. Nothing we used would properly work if the earth were flat.

AS an experienced sailor, and knowing my beloved sailboat, Serenity, very well, I can tell without looking at the knotmeter a one knot difference in speed through the water. And as others who have replied to my original post, looking at the common FE models would require distances that are at least 2 times farther than on a RE, and likely closer to three times farther to circumnavigate. Please believe me that even ignoring GPS data, anyone with enough sailing experience to be doing a roundtheworld race would be aware of the massively increased distance involved. And monohull sailboats have a "hull speed" past which they cannot reasonably go faster. This is based on the wellknown physics of water itself and wave generation. The oversimplified explanation is that as you push a displacement boat hull faster and faster it starts sinking deeper into the water, which greatly increases the drag until you reach a limit = the hull speed. You just cannot make up that degree of distance increase.
My second question/puzzle for FE's is that I have the honor of being old enough to have been trained as a celestial navigator  using a sextant and an accurate watch to determine position at sea. I was lucky enough that when I was in graduate school at Washington University in St. Louis I was able to take a whole semester long course in Celestial taught by a retired Navy officer navigator. Since it was a semester long course we got heavily into the basic math involved. In practice there are books of tables and now calculators/computers that do the math for you, but if you want as a FE believer to assume those books and computers are somehow "doctored" by NASA, my instructor made us do the math ourselves. My final project was to write a computer program that calculated the "great circle route" between two points on the globe. When Lindbergh wanted to calculate the great circle course (the shortest distance) for his historic flight from New York to Paris he had to stretch a string taut on a large globe at the New York City library before the flight and take careful notes as to where he should be at each stage of his flight.
The math required for Celestial is called "spherical trigonometry", and you need a higher level scientific calculator or computer. The simplest example I can give is that on a flat plane an equilateral triangle (all three sides are on the same exact length) has three angles of 60 degrees each. However on a sphere an equilateral triangle is quite different, with three angles of 90 degrees each, not 60. Imagine traveling on a sphere. You start at the north pole and travel due south to the equator. Once you reach the equator you turn due east or west (doesn't matter which), that's a 90 degree turn, and then travel the exact same distance as you did from the pole to the equator. At that point you turn north again (another 90 degree turn) and travel back to the pole. Once back at the pole again you will discover the angle between your departing and returning path is also 90 degrees. Spherical trig has no trouble with calculating all sorts of triangles on the surface of a sphere. Regular flat trig that we all learned in school won't work. Likewise if we really are on a flat earth, then the math behind sextants wouldn't work, not even close. But it works fine. Not as good, and certainly not as easy as GPS, but gets you within about a mile of where you want to be, and for most purposes that's good enough. And it's been used since the invention of a reliable chronometer (accurate clock at sea) in the 1760s. Long before space flight, NASA or any other alleged conspirators. Or do you think all those tens of thousands of crusty old sea captains were all part of a massive conspiracy to hide the flat earth secret for 260 years? And not one of them ever fessed up about the "secret" he/she was keeping?
A FE answer please?

AS an experienced sailor, and knowing my beloved sailboat, Serenity, very well, I can tell without looking at the knotmeter a one knot difference in speed through the water. And as others who have replied to my original post, looking at the common FE models would require distances that are at least 2 times farther than on a RE, and likely closer to three times farther to circumnavigate. Please believe me that even ignoring GPS data, anyone with enough sailing experience to be doing a roundtheworld race would be aware of the massively increased distance involved. And monohull sailboats have a "hull speed" past which they cannot reasonably go faster. This is based on the wellknown physics of water itself and wave generation. The oversimplified explanation is that as you push a displacement boat hull faster and faster it starts sinking deeper into the water, which greatly increases the drag until you reach a limit = the hull speed. You just cannot make up that degree of distance increase.
I've never heard of hull speed, but it is interesting. So basically, if you go TOO fast for a particular vessel (whatever its hull speed is), it will take on water and just sink?

Ships also have a hull speed. Strangely enough the longer the ship is, the higher the hull speed will likely be. I've been on some 1000 foot ships that could go 35 knots, but the fuel consumption was also very high. Basically the hull speed is a function of the hull design and when you start exceeding the hull speed the horsepower (and fuel consumption) starts to increase exponentially. Fast ships have a very long & narrow bow, but when you do that you can carry less cargo but at a faster speed. Docks also charge a fee based upon the length of the vessel. So shorter & wider vessels can carry more cargo with less expensive berthing fees, but you won't get to your next port quite as fast. We would always follow a great circle route that was a nice curve on a flat map, or in our case the electronic map. Sometimes the nice shortest route was followed exactly but if there was a typhoon right in the middle of that path there might be some diversions. Is there any doubt about why? On our trip from Shanghi, China back to California when we reached the most Northern part of the great circle route, usually just South of the Aleutian Islands, we would have a 'top of the world' party for our crew of about 23. If you have 100's of millions of dollars worth of cargo aboard there's little doubt about taking the shortest, fastest route, and that's always the great circle one on the globe earth. If there was another faster route you can be sure that over the last 100 plus years some sailor would have figured it out and would have taken great advantage of that knowledge. If you don't believe that just look up the works of Nathaniel Bowditch. Every American flagged ship still has to carry 'the bowditch' aboard, by law.

The slightly longer but still simplified explanation of hull speed is that the curved bottom of a sailboat can be viewed as an upside down airplane wing. On airplanes the curve points up, on sailboats down. As the boat moves through the water it creates “lift” downwards. The faster it goes the greater the down force, the deeper the boat sinks, and up goes the drag. There’s wave making issues too which limit the speed. And we’re only talking about the boat sinking down by a few inches but that’s enough to increase the drag to limit the speed. The standard formula is that the hull speed of most sailboats in knots is 1.34 times the square root of the waterline length. My 34 foot (Overall) nonracing sailboat has a waterline length of about 25 feet, so her hull speed is 1.34 x sq rt(25) = 6.7 knots. Under perfect conditions I can get her up to about 7 knots but that’s it. (1 knot = 1.15 mph). High end racing yacht designers can use tricks to increase the hull speed past that formula  but not by much. Powerboats generally avoid the problem by skimming (planing) on the surface of the water, like continuously skipping a flat stone. And if the hull is really long and skinny (a ratio of length to beam generally greater than 20:1) other physics kicks in. Note: a real naval architect would be howling about my oversimplification. It ain’t simple.

One other brief comment about sailboats and hull speed. Sailors are warned that if they ever need to be towed by a powerful powerboat  to make absolutely sure the skipper of the powerboat understands the hull speed of the towed boat. Trying to exceed that speed puts tremendous loads on the tow rope and whatever it is attached to, and yes, there have been occasional reports of sailboats being swamped and lost because someone tried to greatly exceed the boat’s hull speed during a tow. They weren’t paying attention as their tow sunk behind them.

AS an experienced sailor, and knowing my beloved sailboat, Serenity, very well, I can tell without looking at the knotmeter a one knot difference in speed through the water. And as others who have replied to my original post, looking at the common FE models would require distances that are at least 2 times farther than on a RE, and likely closer to three times farther to circumnavigate. Please believe me that even ignoring GPS data, anyone with enough sailing experience to be doing a roundtheworld race would be aware of the massively increased distance involved. And monohull sailboats have a "hull speed" past which they cannot reasonably go faster. This is based on the wellknown physics of water itself and wave generation. The oversimplified explanation is that as you push a displacement boat hull faster and faster it starts sinking deeper into the water, which greatly increases the drag until you reach a limit = the hull speed. You just cannot make up that degree of distance increase.
My second question/puzzle for FE's is that I have the honor of being old enough to have been trained as a celestial navigator  using a sextant and an accurate watch to determine position at sea. I was lucky enough that when I was in graduate school at Washington University in St. Louis I was able to take a whole semester long course in Celestial taught by a retired Navy officer navigator. Since it was a semester long course we got heavily into the basic math involved. In practice there are books of tables and now calculators/computers that do the math for you, but if you want as a FE believer to assume those books and computers are somehow "doctored" by NASA, my instructor made us do the math ourselves. My final project was to write a computer program that calculated the "great circle route" between two points on the globe. When Lindbergh wanted to calculate the great circle course (the shortest distance) for his historic flight from New York to Paris he had to stretch a string taut on a large globe at the New York City library before the flight and take careful notes as to where he should be at each stage of his flight.
The math required for Celestial is called "spherical trigonometry", and you need a higher level scientific calculator or computer. The simplest example I can give is that on a flat plane an equilateral triangle (all three sides are on the same exact length) has three angles of 60 degrees each. However on a sphere an equilateral triangle is quite different, with three angles of 90 degrees each, not 60. Imagine traveling on a sphere. You start at the north pole and travel due south to the equator. Once you reach the equator you turn due east or west (doesn't matter which), that's a 90 degree turn, and then travel the exact same distance as you did from the pole to the equator. At that point you turn north again (another 90 degree turn) and travel back to the pole. Once back at the pole again you will discover the angle between your departing and returning path is also 90 degrees. Spherical trig has no trouble with calculating all sorts of triangles on the surface of a sphere. Regular flat trig that we all learned in school won't work. Likewise if we really are on a flat earth, then the math behind sextants wouldn't work, not even close. But it works fine. Not as good, and certainly not as easy as GPS, but gets you within about a mile of where you want to be, and for most purposes that's good enough. And it's been used since the invention of a reliable chronometer (accurate clock at sea) in the 1760s. Long before space flight, NASA or any other alleged conspirators. Or do you think all those tens of thousands of crusty old sea captains were all part of a massive conspiracy to hide the flat earth secret for 260 years? And not one of them ever fessed up about the "secret" he/she was keeping?
A FE answer please?
Regarding your second question , am assuming use of a compass, on FE head due south from the N pole  reach the equator , turn 90 east travel along the equator , which is a circle , following your compass heading of due east for same distance ,turn 90 N travelling same distance again using compass . Wouldn't you end back up at the N pole or am I missing something ?

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole. The math would then fail. On a globe my route from pole to equator and back to the pole is a perfectly fine 909090 equilateral spherical triangle. Yours is a pie wedge.
And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat. Plus the matching issue as to why sea captains don’t believe in a flat earth.

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole. The math would then fail. On a globe my route from pole to equator and back to the pole is a perfectly fine 909090 equilateral spherical triangle. Yours is a pie wedge.
And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat. Plus the matching issue as to why sea captains don’t believe in a flat earth.
They are the same journey with the same result . The math doesn't fail . In FE the angle between outward and return leg at the pole is 90 degrees . The angle at the equator is 90 E and when you reach the required distance you turn 90N . The difference is that you believe you are journeying over a curved surface and so are applying spherical trig.
This also answers your celestial navigation  latitudes are circular lines around the point on earth beneath the N pole.

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole. The math would then fail. On a globe my route from pole to equator and back to the pole is a perfectly fine 909090 equilateral spherical triangle. Yours is a pie wedge.
And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat. Plus the matching issue as to why sea captains don’t believe in a flat earth.
Everything above our heads appears as if it is in a cylinder.
That is how objects would appear above an x/y coordinate system.
You don't need spherical trigonometry to negotiate an ocean trip.
That's just nonsense.

Okay, I gave the illustration of an equilateral triangle on a sphere with three 90 degree angles to contrast to the standard flat equilateral triangle of three 60 degree angles just to illustrate the difference and to make sure readers knew what I was talking about. Nevertheless spherical trig is indeed different than flat trig, and celestial navigation is based on spherical trig. It uses spherical trig to calculate the angles relative to where a celestial body is directly over the surface of the earth at the time you are taking the sight. (In other words the place you would be if the celestial body you are using was straight up over your head.) If that were the case all the time then celestial would be absurdly easy. Since the celestial object (sun, moon stars, planets) is almost never directly over your head, you use a sextant to measure the angle that object appears above the horizon. Celestial then uses that angle with spherical trig to determine your "line of position". (It's a movie myth that a single sextant sight gives you an "X marks the spot" position. Except for the famous "noon sight" you get a short line that you are somewhere on that line. There are other ways to then upscale that to a more precise position.). And in reality nobody does the math themselves. You use either books of detailed tables or later computer/calculators to do that math. BUT the reality is that the system is indeed based on spherical trig, not flat trig. The respondent's statement that you don't need spherical trig to navigate, that it's "nonsense" is simply not true for celestial navigation. it is true for navigating on a chart, within sight of land, to work out your bearing to a lighthouse for example. Flat trig works fine for that, because the distances involved are so short that the "needed correction" for being on the surface of a sphere is trivial. It's the same reason you can use a flat paper chart to navigate along the coast. Technically that flat chart is slightly "wrong" since you cannot accurately depict a sphere on a flat paper, but again for the distances involved the "error" is trivial. But out on the open ocean, trying to determine your position is an entirely different matter. Prior to GPS, you needed celestial navigation to do that. And without the spherical trig behind celestial, it would fail. And it doesn't fail  it works. It's worked for hundreds of years. If the earth was flat, then the math for celestial would be standard flat trig, but it's not. There would be no need to calculate spherical triangles to determine your position since you are on a alleged flat surface. This is a simple nondisputable fact.
I feel I speak with some degree of realworld expertise about celestial navigation. Anybody out there who has used celestial who disagrees with me?

Okay, I gave the illustration of an equilateral triangle on a sphere with three 90 degree angles to contrast to the standard flat equilateral triangle of three 60 degree angles just to illustrate the difference and to make sure readers knew what I was talking about. Nevertheless spherical trig is indeed different than flat trig, and celestial navigation is based on spherical trig. It uses spherical trig to calculate the angles relative to where a celestial body is directly over the surface of the earth at the time you are taking the sight. (In other words the place you would be if the celestial body you are using was straight up over your head.) If that were the case all the time then celestial would be absurdly easy. Since the celestial object (sun, moon stars, planets) is almost never directly over your head, you use a sextant to measure the angle that object appears above the horizon. Celestial then uses that angle with spherical trig to determine your "line of position". (It's a movie myth that a single sextant sight gives you an "X marks the spot" position. Except for the famous "noon sight" you get a short line that you are somewhere on that line. There are other ways to then upscale that to a more precise position.). And in reality nobody does the math themselves. You use either books of detailed tables or later computer/calculators to do that math. BUT the reality is that the system is indeed based on spherical trig, not flat trig. The respondent's statement that you don't need spherical trig to navigate, that it's "nonsense" is simply not true for celestial navigation. it is true for navigating on a chart, within sight of land, to work out your bearing to a lighthouse for example. Flat trig works fine for that, because the distances involved are so short that the "needed correction" for being on the surface of a sphere is trivial. It's the same reason you can use a flat paper chart to navigate along the coast. Technically that flat chart is slightly "wrong" since you cannot accurately depict a sphere on a flat paper, but again for the distances involved the "error" is trivial. But out on the open ocean, trying to determine your position is an entirely different matter. Prior to GPS, you needed celestial navigation to do that. And without the spherical trig behind celestial, it would fail. And it doesn't fail  it works. It's worked for hundreds of years. If the earth was flat, then the math for celestial would be standard flat trig, but it's not. There would be no need to calculate spherical triangles to determine your position since you are on a alleged flat surface. This is a simple nondisputable fact.
I feel I speak with some degree of realworld expertise about celestial navigation. Anybody out there who has used celestial who disagrees with me?
Since all things above our heads on an x/y plane would appear as if they were moving in a cylindrical motion, that accounts for your supposed spherical trig...but like I wrote earlier, it isn't even necessary.
Just a bunch of hogwash.

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole. The math would then fail. On a globe my route from pole to equator and back to the pole is a perfectly fine 909090 equilateral spherical triangle. Yours is a pie wedge.
And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat. Plus the matching issue as to why sea captains don’t believe in a flat earth.
Everything above our heads appears as if it is in a cylinder.
That is how objects would appear above an x/y coordinate system.
You don't need spherical trigonometry to negotiate an ocean trip.
That's just nonsense.
A sextant uses the angle of celestial objects to the horizon in a formula to plot it onto a sphere, and they have been used for centuries and been proven accurate. It matches up perfectly with GPS coordinates as stated before, they are both commonly used together even now.
Can you please explain the math behind your cylindrical x/y coordinate system? How do you use that to navigate? Thanks.

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole. The math would then fail. On a globe my route from pole to equator and back to the pole is a perfectly fine 909090 equilateral spherical triangle. Yours is a pie wedge.
And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat. Plus the matching issue as to why sea captains don’t believe in a flat earth.
Everything above our heads appears as if it is in a cylinder.
That is how objects would appear above an x/y coordinate system.
You don't need spherical trigonometry to negotiate an ocean trip.
That's just nonsense.
A sextant uses the angle of celestial objects to the horizon in a formula to plot it onto a sphere, and they have been used for centuries and been proven accurate. It matches up perfectly with GPS coordinates as stated before, they are both commonly used together even now.
Can you please explain the math behind your cylindrical x/y coordinate system? How do you use that to navigate? Thanks.
A sextant is not used to plot things onto a sphere.
Where did you get that from?
They do not carry globes on ships.
They carry flat charts.

A sextant uses the angle of celestial objects to the horizon in a formula to plot it onto a sphere, and they have been used for centuries and been proven accurate. It matches up perfectly with GPS coordinates as stated before, they are both commonly used together even now.
Can you please explain the math behind your cylindrical x/y coordinate system? How do you use that to navigate? Thanks.
A sextant is not used to plot things onto a sphere.
Where did you get that from?
They do not carry globes on ships.
They carry flat charts.
Sextants do indeed plot onto a sphere. I got that from the manual of the sextant I own, plus every article describing how they work on the internet.
Here is one  https://www.lloydm.net/Celestial/Navigation.html
Part 2  Applying Spherical Trigonometry. In this section the two formulas of spherical trigonometry that are needed to solve the celestial navigation problem will be derived. Recall from Part 1 that a great circle is any circle on the surface of a sphere whose center is also the center of the sphere. The angle of intersection of two arcs of great circles is defined as the angle formed by the tangents to these arcs at their point of intersection.
Page 33 of this document has some images explaining how to use a sextants measurements to plot onto a sphere  https://pbps.org/docs/Celestial%20Navigation%20Book.pdf
Yes, ships don't carry 100's of huge globes with them because there isn't room for them. Flat maps take up much less space and are easier to draw and measure on.

Okay, I gave the illustration of an equilateral triangle on a sphere with three 90 degree angles to contrast to the standard flat equilateral triangle of three 60 degree angles just to illustrate the difference and to make sure readers knew what I was talking about. Nevertheless spherical trig is indeed different than flat trig, and celestial navigation is based on spherical trig. It uses spherical trig to calculate the angles relative to where a celestial body is directly over the surface of the earth at the time you are taking the sight. (In other words the place you would be if the celestial body you are using was straight up over your head.) If that were the case all the time then celestial would be absurdly easy. Since the celestial object (sun, moon stars, planets) is almost never directly over your head, you use a sextant to measure the angle that object appears above the horizon. Celestial then uses that angle with spherical trig to determine your "line of position". (It's a movie myth that a single sextant sight gives you an "X marks the spot" position. Except for the famous "noon sight" you get a short line that you are somewhere on that line. There are other ways to then upscale that to a more precise position.). And in reality nobody does the math themselves. You use either books of detailed tables or later computer/calculators to do that math. BUT the reality is that the system is indeed based on spherical trig, not flat trig. The respondent's statement that you don't need spherical trig to navigate, that it's "nonsense" is simply not true for celestial navigation. it is true for navigating on a chart, within sight of land, to work out your bearing to a lighthouse for example. Flat trig works fine for that, because the distances involved are so short that the "needed correction" for being on the surface of a sphere is trivial. It's the same reason you can use a flat paper chart to navigate along the coast. Technically that flat chart is slightly "wrong" since you cannot accurately depict a sphere on a flat paper, but again for the distances involved the "error" is trivial. But out on the open ocean, trying to determine your position is an entirely different matter. Prior to GPS, you needed celestial navigation to do that. And without the spherical trig behind celestial, it would fail. And it doesn't fail  it works. It's worked for hundreds of years. If the earth was flat, then the math for celestial would be standard flat trig, but it's not. There would be no need to calculate spherical triangles to determine your position since you are on a alleged flat surface. This is a simple nondisputable fact.
I feel I speak with some degree of realworld expertise about celestial navigation. Anybody out there who has used celestial who disagrees with me?
Celestial navigation has been carried out for thousands of years. Mankind knows the celestial sphere rotates overhead and has always used this for navigation .The stars rise and set , and all reach zenith which has always given a good approximation of longitude .
The spherical calculations are pointless as shown , none of what you say points to earth being a globe .
A trip from N pole to the equator and back after travelling a quarter of the equators length is not a short journey so how can you say it only works on short trips?

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole. The math would then fail. On a globe my route from pole to equator and back to the pole is a perfectly fine 909090 equilateral spherical triangle. Yours is a pie wedge.
And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat. Plus the matching issue as to why sea captains don’t believe in a flat earth.
Everything above our heads appears as if it is in a cylinder.
Only if you consider the southern hemisphere doesn't exist.

It doesn't.

It doesn't.
Very clever.
Why is there "another cylinder" south of the Equator?
Anyway, what's so special about the Equator on a flat Earth? What's so special about the tropics on a flat Earth?

Not clever . It's a statement of the bleedin obvious.
The tropics mark the Northern and Southernmost limits of the sun's journey as it orbits around the plane . The equator is the midpoint on it's journey .
The globe model has the tropics at 66.6 degrees from the poles and the polar circles at 66.6 degrees from the equator . The apparent cause is the tilt of the earths axis being 23.4 degrees  which is 66.6 degrees if you take the angle from the horizontal.
And earth scoots around the sun at an average distance of 93,000,000 miles and completes it orbit in 365.25 days (hence our need for leap years) which gives a velocity of 66,600 mph.
Interesting that.

Sextants do indeed plot onto a sphere. I got that from the manual of the sextant I own, plus every article describing how they work on the internet.
Here is one  https://www.lloydm.net/Celestial/Navigation.html
Part 2  Applying Spherical Trigonometry. In this section the two formulas of spherical trigonometry that are needed to solve the celestial navigation problem will be derived. Recall from Part 1 that a great circle is any circle on the surface of a sphere whose center is also the center of the sphere. The angle of intersection of two arcs of great circles is defined as the angle formed by the tangents to these arcs at their point of intersection.
Page 33 of this document has some images explaining how to use a sextants measurements to plot onto a sphere  https://pbps.org/docs/Celestial%20Navigation%20Book.pdf
Yes, ships don't carry 100's of huge globes with them because there isn't room for them. Flat maps take up much less space and are easier to draw and measure on.
I am just going to boil things down in your reply...
"I know I claimed that sextants plot on a sphere; because, while I do understand ships do not carry around a globe for the purpose of drawing lines on it (...they carry perfectly flat charts where they end up plotting what they derive from the sextant, just like you wrote totallackey!), I gonna stick with what the writing says because it fits my narrative. Real life experience be damned."
Yeah...ain't buyin' it...
But thanks for admitting what I wrote was true.
There is hope for you yet.

Sextants do indeed plot onto a sphere. I got that from the manual of the sextant I own, plus every article describing how they work on the internet.
Here is one  https://www.lloydm.net/Celestial/Navigation.html
Part 2  Applying Spherical Trigonometry. In this section the two formulas of spherical trigonometry that are needed to solve the celestial navigation problem will be derived. Recall from Part 1 that a great circle is any circle on the surface of a sphere whose center is also the center of the sphere. The angle of intersection of two arcs of great circles is defined as the angle formed by the tangents to these arcs at their point of intersection.
Page 33 of this document has some images explaining how to use a sextants measurements to plot onto a sphere  https://pbps.org/docs/Celestial%20Navigation%20Book.pdf
Yes, ships don't carry 100's of huge globes with them because there isn't room for them. Flat maps take up much less space and are easier to draw and measure on.
I am just going to boil things down in your reply...
"I know I claimed that sextants plot on a sphere; because, while I do understand ships do not carry around a globe for the purpose of drawing lines on it (...they carry perfectly flat charts where they end up plotting what they derive from the sextant, just like you wrote totallackey!), I gonna stick with what the writing says because it fits my narrative. Real life experience be damned."
Yeah...ain't buyin' it...
But thanks for admitting what I wrote was true.
There is hope for you yet.
You claimed sextants don't plot onto a sphere, I showed that yes, the math indeed takes a sextants measurements and plots it onto a sphere. This is a simple fact, the math is very clear.
Putting words in my mouth doesn't prove otherwise, it's rude and doesn't contribute to the discussion, so please stop doing that.
Can you show me what math you think sextants use that don't require spherical mapping and calculations?

You claimed sextants don't plot onto a sphere, I showed that yes, the math indeed takes a sextants measurements and plots it onto a sphere. This is a simple fact, the math is very clear.
You showed that math doesn't actually plot things on a sphere.
You admitted the plotting is circumscribed on flat charts found on ships.
That too is a simple fact.
Putting words in my mouth doesn't prove otherwise, it's rude and doesn't contribute to the discussion, so please stop doing that.
I objectively paraphrased what you wrote.
Nothing wrong with that at all.
Can you show me what math you think sextants use that don't require spherical mapping and calculations?
Obviously none of the math is based on spherical mapping and calculations because the plotting is done on a flat chart.

You claimed sextants don't plot onto a sphere, I showed that yes, the math indeed takes a sextants measurements and plots it onto a sphere. This is a simple fact, the math is very clear.
You showed that math doesn't actually plot things on a sphere.
Please show where I showed math that doesn't plot data onto a sphere for a sextant. I'll quote myself earlier, saying the opposite.
Here is one  https://www.lloydm.net/Celestial/Navigation.html
Part 2  Applying Spherical Trigonometry. In this section the two formulas of spherical trigonometry that are needed to solve the celestial navigation problem will be derived. Recall from Part 1 that a great circle is any circle on the surface of a sphere whose center is also the center of the sphere. The angle of intersection of two arcs of great circles is defined as the angle formed by the tangents to these arcs at their point of intersection.
Obviously none of the math is based on spherical mapping and calculations because the plotting is done on a flat chart.
If I draw a picture of you on paper does it mean you are flat?

You claimed sextants don't plot onto a sphere, I showed that yes, the math indeed takes a sextants measurements and plots it onto a sphere. This is a simple fact, the math is very clear.
You showed that math doesn't actually plot things on a sphere.
Please show where I showed math that doesn't plot data onto a sphere for a sextant. I'll quote myself earlier, saying the opposite.
You stated quite succinctly that ships do not carry spheres for charting.
You stated they carry flat charts and maps.
You wrote it, I didn't.
Why would you ask me to show you something you wrote?
If I draw a picture of you on paper does it mean you are flat?
You and I both know that I am not flat.
The reason we know this is simple.
Simply by standing in the same room, you can see that.
No such luck with the supposed globe.

The use of a sextant isn't a complicated matter JSS .
Sextants measure angles up to 60 degrees . Used at sea to measure the angle between the horizon and whichever celestial body you want to use for navigation .
On land you you would set up a sextant to measure the angle between the horizontal , found by plumb line and set square , and which ever body . You don't use a plumb bob at sea because it would be difficult to set .
If you want to measure an angle greater than 60 degrees you use a quadrant .
If you want to plot your results onto a globe you would apply spherical trig .
It's just a device for measuring angles between points and says nothing about the earth being a sphere .

The use of a sextant isn't a complicated matter JSS .
Sextants measure angles up to 60 degrees . Used at sea to measure the angle between the horizon and whichever celestial body you want to use for navigation .
On land you you would set up a sextant to measure the angle between the horizontal , found by plumb line and set square , and which ever body . You don't use a plumb bob at sea because it would be difficult to set .
If you want to measure an angle greater than 60 degrees you use a quadrant .
If you want to plot your results onto a globe you would apply spherical trig .
It's just a device for measuring angles between points and says nothing about the earth being a sphere .
If plotting your position has nothing to do with spheres then why use spherical trig?

To map your findings onto a sphere .

You claimed sextants don't plot onto a sphere, I showed that yes, the math indeed takes a sextants measurements and plots it onto a sphere. This is a simple fact, the math is very clear.
You showed that math doesn't actually plot things on a sphere.
Please show where I showed math that doesn't plot data onto a sphere for a sextant. I'll quote myself earlier, saying the opposite.
You stated quite succinctly that ships do not carry spheres for charting.
You stated they carry flat charts and maps.
You wrote it, I didn't.
I think I see what you mean now. You are claiming that because I said ships don't carry globes, the math for sextants doesn't use spherical geometry?
That is deeply flawed logic. What ships carry has nothing to do with math for calculating a position with a sextant.
The math for sextants uses spherical calculations, I've shown you this several times. Sextants plot onto a sphere. What you do with that data later doesn't change the math.

To map your findings onto a sphere .
Exactly. You take a reading of an angle with a sextant. You then use math to plot this onto a sphere to find your position.
If you try and map the angle directly it to a plane, it doesn't work.
Thus, the shape of the earth is a sphere.

To map your findings onto a sphere .
Exactly. You take a reading of an angle with a sextant. You then use math to plot this onto a sphere to find your position.
If you try and map the angle directly it to a plane, it doesn't work.
Thus, the shape of the earth is a sphere.
You are now contradicting yourself.
You wrote earlier that the angles are plotted onto flat charts.
Even my great uncle, who used a sextant in WWII in the Pacific Theater, showed me a sextant and showed me to use it.
He drew all the angles on a lfat piece of paper.
He didn't use a sphere.

To map your findings onto a sphere .
Exactly. You take a reading of an angle with a sextant. You then use math to plot this onto a sphere to find your position.
If you try and map the angle directly it to a plane, it doesn't work.
Thus, the shape of the earth is a sphere.
You are now contradicting yourself.
You wrote earlier that the angles are plotted onto flat charts.
Even my great uncle, who used a sextant in WWII in the Pacific Theater, showed me a sextant and showed me to use it.
He drew all the angles on a lfat piece of paper.
He didn't use a sphere.
I think you are getting confused about what plot onto a sphere means. Let me try and be more precise in my wording.
You use a sextant to measure the angles of the heavenly bodies you are measuring.
You take these angles and use a formula to mathematically transform them into a set of spherical coordinates.
You can then take these spherical coordinates and use a pen to draw them onto a physical map, or a globe, or any other surface you can reach.

To map your findings onto a sphere .
Exactly. You take a reading of an angle with a sextant. You then use math to plot this onto a sphere to find your position.
If you try and map the angle directly it to a plane, it doesn't work.
Thus, the shape of the earth is a sphere.
You are now contradicting yourself.
You wrote earlier that the angles are plotted onto flat charts.
Even my great uncle, who used a sextant in WWII in the Pacific Theater, showed me a sextant and showed me to use it.
He drew all the angles on a flat piece of paper.
He didn't use a sphere.
I think you are getting confused about what plot onto a sphere means. Let me try and be more precise in my wording.
You use a sextant to measure the angles of the heavenly bodies you are measuring.
You take these angles and use a formula to mathematically transform them into a set of spherical coordinates.
You can then take these spherical coordinates and use a pen to draw them onto a physical map, or a globe, or any other surface you can reach.
But you observed the celestial sphere (i.e., the cylinder above you).
When you plot on a piece of paper, you are not plotting on a sphere.

I think you are getting confused about what plot onto a sphere means. Let me try and be more precise in my wording.
1. You use a sextant to measure the angles of the heavenly bodies you are measuring.
2. You take these angles and use a formula to mathematically transform them into a set of spherical coordinates.
3. You can then take these spherical coordinates and use a pen to draw them onto a physical map, or a globe, or any other surface you can reach.
But you observed the celestial sphere (i.e., the cylinder above you).
When you plot on a piece of paper, you are not plotting on a sphere.
A sphere is not a cylinder.
Please carefully read what I wrote. I added numbers to refer to the steps easier.
Step 2 is done with math, it transforms the angle onto a sphere. With numbers. Just a calculator.
Step 3 is where you take the result and draw it on something
You are thinking you go from step 1 to step 3, but that is missing step 2.

I think you are getting confused about what plot onto a sphere means. Let me try and be more precise in my wording.
1. You use a sextant to measure the angles of the heavenly bodies you are measuring.
2. You take these angles and use a formula to mathematically transform them into a set of spherical coordinates.
3. You can then take these spherical coordinates and use a pen to draw them onto a physical map, or a globe, or any other surface you can reach.
But you observed the celestial sphere (i.e., the cylinder above you).
When you plot on a piece of paper, you are not plotting on a sphere.
A sphere is not a cylinder.
Please carefully read what I wrote. I added numbers to refer to the steps easier.
Step 2 is done with math, it transforms the angle onto a sphere. With numbers. Just a calculator.
Step 3 is where you take the result and draw it on something
You are thinking you go from step 1 to step 3, but that is missing step 2.
Everything above us appears as if it is in a cylinder, as it would above any x/y coordinate plane.
That is indisputable math.
Period.
The reason for this is that what we see above us is plotted onto a sphere or dome.

Everything above us appears as if it is in a cylinder, as it would above any x/y coordinate plane.
That is indisputable math.
Period.
The reason for this is that what we see above us is plotted onto a sphere or dome.
Going back full circle, a sextant's ability to determine your position on a globe, using a spherical celestial reference would indicate that the heavens are in fact, not a cylinder.
I'm pretty sure I've asked you this before, but how do you plot a sextants measurements onto this x/y cylinder?
Please show this indisputable math.

Everything above us appears as if it is in a cylinder, as it would above any x/y coordinate plane.
That is indisputable math.
Period.
The reason for this is that what we see above us is plotted onto a sphere or dome.
Going back full circle, a sextant's ability to determine your position on a globe, using a spherical celestial reference would indicate that the heavens are in fact, not a cylinder.
I'm pretty sure I've asked you this before, but how do you plot a sextants measurements onto this x/y cylinder?
Please show this indisputable math.
A sextant has no ability to determine position on a globe . It used for measuring an angle . It's basic geometry and that is what it does .
Applying spherical trig is not the sextants ability. Applying spherical maths allows people to plot measured angles on a globe representation of earth irrespective of the fact that it is not a globe.You could do the same with any shape . Doing so converts what the sextant measures into a mathematical model which has no basis in reality .

Everything above us appears as if it is in a cylinder, as it would above any x/y coordinate plane.
That is indisputable math.
Period.
The reason for this is that what we see above us is plotted onto a sphere or dome.
Going back full circle, a sextant's ability to determine your position on a globe, using a spherical celestial reference would indicate that the heavens are in fact, not a cylinder.
I'm pretty sure I've asked you this before, but how do you plot a sextants measurements onto this x/y cylinder?
Please show this indisputable math.
A sextant has no ability to determine position on a globe . It used for measuring an angle . It's basic geometry and that is what it does .
Applying spherical trig is not the sextants ability. Applying spherical maths allows people to plot measured angles on a globe representation of earth irrespective of the fact that it is not a globe.You could do the same with any shape . Doing so converts what the sextant measures into a mathematical model which has no basis in reality .
A sextant can't "apply spherical trig" any more than a ruler or a potato can, it's just a tool. It gives you the numbers.
And you can not in fact, do it on any shape. The math for turning sextant angles into coordinates uses spherical math and that does not work on any other geometric shape. If it worked on a plane, it would be called planar math. This is not the case no matter how many times you say it.
You can claim it has no basis in reality, but every sailor would disagree. They work, they have worked for hundreds of years.

Okay, I'm the one who started this. This will be my final input. But I'd like to make a few last points/clarifications:
1) I guess in the beginning I should have defined terms. What I mean my "celestial navigation" is the final product of years of smart people working on this problem of using a sextant, a chronometer, and carefully developed tables or computers/calculators to find your position at sea. The math required to do so is spherical trigonometry. This is a fact. Spherical trig is used because you are finding your position on a sphere. So, my definition of "celestial navigation" is the common system in worldwide use before the advent of electronic navigation systems like GPS, LORAN, DECCA, etc.
2) It is true that people have been using other kinds of celestial navigation for millennia. The indigenous navigators of the South Pacific are particularly famous. (Read the book about them: "We, the Navigators" by David Lewis  fascinating!) Even in more modern ages there is a "lunar method" that does not require a chronometer. The famous sea captain, Joshua Slocum, who was the first person to sail singlehanded around the world in 189598 reportedly used that method, but I understand it is extraordinarily complicated. However Slocum spent years before that famous voyage as a regular sea captain, so he had the experience & knowledge.
3) The example I gave of a large spherical triangle based on the pole and the equator was just that, an example. One easily imagined in someone's mind. In real life celestial navigation involves solving many different spherical triangles, some big, some small. There are an infinite number of possible spherical triangles on the surface of a sphere, just like there are an infinite number of potential flat triangles on a plane. I was just giving an example. Don't get hung up on that.
4) It's also true that navigators use flat maps (charts) at sea, not a globe. BUT, BUT, BUT that's because flat maps are much more convenient, AND over the scale of the typical marine chart the differences between a flat Mercator projection chart and the real spherical world are trivial. But that is for the same reason that when you are standing on shore looking at the horizon  it looks flat. It's NOT FLAT! But the curvature is so subtle that the human eye and brain cannot detect it. Similarly over the distances covered by a typical chart "correcting" for the fact that this is a flat representation of a small portion of a sphere is just unneeded. BUT for crossing entire oceans, then it matters. You then use a great circle course, which when plotted on a standard Mercator projection chart ends up looking like a "longer" curved line, but in the real world is the shortest distance. The ONLY great circle routes when plotted on a Mercator chart are still straight lines are at the Equator headed due east or west or when traveling absolutely due north/south along a line of longitude. All other great circle routes on a standard chart will look like a curved line. In an earlier post I mentioned Charles Lindbergh. Before his historic New York to Paris flight, he went to the main New York City public library and stretched a string on a very large globe they had there (I understand it was 45 feet in diameter) between New York and Paris. Since he could not take a globe with him in his airplane, he took careful notes as to where he needed to be for each stage of his flight to follow a great circle course. And in almost every book I've read about Lindbergh and his flight, there is a chart of his course across the Atlantic plotted on a Mercator projection chart of the Atlantic ocean. The course looks like a curved line. In reality, up in the air, it was not  it was a straightashecoulddoit straight Great Circle line to Paris.
4) And again, all those crusty old sea captains who had to do celestial never fessed up? And none of them ever found the "ice wall" circling the flat earth? And since the southern oceans were at one point heavily sailed by whalers (thankfully that's over) you think one of them would have noticed an endless wall of ice?
5) And I need at least mention another inconvenient fact about celestial that I haven't yet mentioned. Celestial is also based heavily on a bunch of smart astronomers figuring out precisely where the "geographic point" is for all the celestial bodies used for navigation are at any given time. The celestial bodies are: sun, moon, major planets, and about 20 of the brightest stars. The principle is using the angle measured by a sextant and a chronometer to calculate where you are compared to the geographic point of the celestial object you are using at that exact point in time. (Given the rotation of the Earth, for every 4 seconds you are off on your time measurement your calculated position is off by a mile of longitude.) The geographic point is where the celestial object would be directly over your head. If the object is low on the horizon from the navigator's perspective the geographic point could be thousands of miles away. That's not a problem, but then the use of spherical trig is mandatory or else the error involved would make the effort pointless. Plus, that entire process of calculating and then incorporating the geographic point for all those celestial bodies into the tables or calculators/computers used is also based on the earth being a globe. None of those astronomers are flatearthers. So geographic points are yet another critical part of celestial navigation that relies on the Earth being a sphere.
6) To the FE people who have replied, it strikes me that none have you have carefully thought about what I and others have said about celestial navigation. Your response seems to be the intellectual equivalent of putting your fingers in your ears and yelling "NaNaNaNa" when someone is telling you something you don't want to hear.
7) And no FE responder has explained why the distances required for an aroundtheworld sailing race would be 23 times longer based on the common FE model than they really are. And, yes, the sailors would notice that sort of thing.
8) Finally  Celestial navigation works. It's worked for centuries. Long before NASA and at least since the 1760s by every sea captain from every seafaring nation, regardless of politics, religion. The math required is spherical trigonometry. It's spherical trigonometry because the earth is a sphere. No one would bother using the complexity of spherical trig if it wasn't needed. Be intellectually honest and get over it. I'm going sailing.

Everything above us appears as if it is in a cylinder, as it would above any x/y coordinate plane.
That is indisputable math.
Period.
It simply doesn't.
Just look at star trails showing the celestial equator: https://earthsky.org/todaysimage/compositephotostartrailsshowcelestialequator . How do you get that from a cylinder?
Star trails pictures are something anyone can do with fairly basic equipment, maybe not a lowend pointandshoot or a smartphone but nothing exceptional, just a decent DSLR and a tripod. If you don't live very close to the poles you'll be able to see the celestial equator.

Everything above us appears as if it is in a cylinder, as it would above any x/y coordinate plane.
That is indisputable math.
Period.
It simply doesn't.
Just look at star trails showing the celestial equator: https://earthsky.org/todaysimage/compositephotostartrailsshowcelestialequator . How do you get that from a cylinder?
Star trails pictures are something anyone can do with fairly basic equipment, maybe not a lowend pointandshoot or a smartphone but nothing exceptional, just a decent DSLR and a tripod. If you don't live very close to the poles you'll be able to see the celestial equator.
Take it up with your local mathematician and your textbooks.

Okay, I'm the one who started this. This will be my final input. But I'd like to make a few last points/clarifications:
1) I guess in the beginning I should have defined terms. What I mean my "celestial navigation" is the final product of years of smart people working on this problem of using a sextant, a chronometer, and carefully developed tables or computers/calculators to find your position at sea. The math required to do so is spherical trigonometry. This is a fact. Spherical trig is used because you are finding your position on a sphere. So, my definition of "celestial navigation" is the common system in worldwide use before the advent of electronic navigation systems like GPS, LORAN, DECCA, etc.
Spherical trig is used because of the celestial sphere.
2) It is true that people have been using other kinds of celestial navigation for millennia. The indigenous navigators of the South Pacific are particularly famous. (Read the book about them: "We, the Navigators" by David Lewis  fascinating!) Even in more modern ages there is a "lunar method" that does not require a chronometer. The famous sea captain, Joshua Slocum, who was the first person to sail singlehanded around the world in 189598 reportedly used that method, but I understand it is extraordinarily complicated. However Slocum spent years before that famous voyage as a regular sea captain, so he had the experience & knowledge.
An entire additional paragraph that actually states little...
3) The example I gave of a large spherical triangle based on the pole and the equator was just that, an example. One easily imagined in someone's mind. In real life celestial navigation involves solving many different spherical triangles, some big, some small. There are an infinite number of possible spherical triangles on the surface of a sphere, just like there are an infinite number of potential flat triangles on a plane. I was just giving an example. Don't get hung up on that.
You mean hung up on all these spherical triangles that don't exist...
Don't worry...
4) It's also true that navigators use flat maps (charts) at sea, not a globe. BUT, BUT, BUT that's because flat maps are much more convenient, AND over the scale of the typical marine chart the differences between a flat Mercator projection chart and the real spherical world are trivial. But that is for the same reason that when you are standing on shore looking at the horizon  it looks flat. It's NOT FLAT! But the curvature is so subtle that the human eye and brain cannot detect it. Similarly over the distances covered by a typical chart "correcting" for the fact that this is a flat representation of a small portion of a sphere is just unneeded. BUT for crossing entire oceans, then it matters. You then use a great circle course, which when plotted on a standard Mercator projection chart ends up looking like a "longer" curved line, but in the real world is the shortest distance. The ONLY great circle routes when plotted on a Mercator chart are still straight lines are at the Equator headed due east or west or when traveling absolutely due north/south along a line of longitude. All other great circle routes on a standard chart will look like a curved line. In an earlier post I mentioned Charles Lindbergh. Before his historic New York to Paris flight, he went to the main New York City public library and stretched a string on a very large globe they had there (I understand it was 45 feet in diameter) between New York and Paris. Since he could not take a globe with him in his airplane, he took careful notes as to where he needed to be for each stage of his flight to follow a great circle course. And in almost every book I've read about Lindbergh and his flight, there is a chart of his course across the Atlantic plotted on a Mercator projection chart of the Atlantic ocean. The course looks like a curved line. In reality, up in the air, it was not  it was a straightashecoulddoit straight Great Circle line to Paris.
Those great circle lines you write about are the translations of the celestial sphere to the flat earth chart.
4) And again, all those crusty old sea captains who had to do celestial never fessed up? And none of them ever found the "ice wall" circling the flat earth? And since the southern oceans were at one point heavily sailed by whalers (thankfully that's over) you think one of them would have noticed an endless wall of ice?
Captain Cook  "The ice extended east and west, far beyond the reach of our sight while the southern half of the horizon was illuminated by rays of light which were reflected by the ice to a considerable height."
5) And I need at least mention another inconvenient fact about celestial that I haven't yet mentioned. Celestial is also based heavily on a bunch of smart astronomers figuring out precisely where the "geographic point" is for all the celestial bodies used for navigation are at any given time. The celestial bodies are: sun, moon, major planets, and about 20 of the brightest stars. The principle is using the angle measured by a sextant and a chronometer to calculate where you are compared to the geographic point of the celestial object you are using at that exact point in time. (Given the rotation of the Earth, for every 4 seconds you are off on your time measurement your calculated position is off by a mile of longitude.) The geographic point is where the celestial object would be directly over your head. If the object is low on the horizon from the navigator's perspective the geographic point could be thousands of miles away. That's not a problem, but then the use of spherical trig is mandatory or else the error involved would make the effort pointless. Plus, that entire process of calculating and then incorporating the geographic point for all those celestial bodies into the tables or calculators/computers used is also based on the earth being a globe. None of those astronomers are flatearthers. So geographic points are yet another critical part of celestial navigation that relies on the Earth being a sphere.
Who were the smart astronomers when they first started doing it?
Got some names?
6) To the FE people who have replied, it strikes me that none have you have carefully thought about what I and others have said about celestial navigation. Your response seems to be the intellectual equivalent of putting your fingers in your ears and yelling "NaNaNaNa" when someone is telling you something you don't want to hear.
Actually, it seems that way to any person when their argument is blown to bits.
7) And no FE responder has explained why the distances required for an aroundtheworld sailing race would be 23 times longer based on the common FE model than they really are. And, yes, the sailors would notice that sort of thing.
Maybe cause they wouldn't be 23 times longer?
8) Finally  Celestial navigation works. It's worked for centuries. Long before NASA and at least since the 1760s by every sea captain from every seafaring nation, regardless of politics, religion. The math required is spherical trigonometry. It's spherical trigonometry because the earth is a sphere. No one would bother using the complexity of spherical trig if it wasn't needed. Be intellectually honest and get over it. I'm going sailing.
It's spherical trigonometry based on celestial sphere, translated to a flat earth chart.
Just like reality.

7) And no FE responder has explained why the distances required for an aroundtheworld sailing race would be 23 times longer based on the common FE model than they really are. And, yes, the sailors would notice that sort of thing.
Maybe cause they wouldn't be 23 times longer?
I am not familiar with the details of navigation and boating, and I tend to not post on anything unless I can personally grasp it. And this is something I can certainly comprehend. Just a modicum of spatial reasoning can show it with a thought experiment:
Imagine the globe model of the earth that shows lines of latitude. Imagine looking at the north pole. Notice how all the latitudes get smaller the further north they go? Now, turn it over and look at the south pole where Antarctica is shown. Again, the the latitudes get smaller as they approach the pole, each one further south (instead of north, this time).
This means that if the globe model is accurate, circumnavigating the planet along the Tropic of Capricorn (the latitude halfway between the equator and S. pole) and the Tropic of Cancer (the latitude between equator and N. pole) is the same distance (yes, I know that landmasses are in the way, pretend they're not for this thought experiment, or that you travel overland for those portions).
However, on the FE models as depicted on this website, the Tropic of Capricorn would take a MUCH longer time to travel around to meet back where you started. It's a far bigger circle  hence it would take a lot longer going the same speed.
Hence, BoatBum's comment. The problem that the FE model needs to wrestle with and provide a satisfactory answer for is that sailors don't report significantly longer travel times in the southern hemisphere.
One possible answer to this that have been brought up in this thread is that ships are simply traveling much faster, and the sailors don't know how fast they're going. But that idea has been responded to in great detail as virtually impossible and impractical  sailors do know their speed.

Ah, I love selective quotes. Total Lackey is quoting from the accounts of Cook's second voyage. Please read the entire section. He was referring to what was called then "field ice" = floating pieces of ice of varying sizes, nowadays called pack ice. Cook went on to lament that the ice prevented him from sailing further south, and he speculated that beyond the pack ice lay a land which contained the notyetdiscovered South Pole. Keep in mind that from the deck of a ship all you can see is to about 8 miles away under perfect clear sky conditions. After that it's over the horizon. So he turned around and headed north. Thus is all from page 160 of the accounts of Cook's second voyage. Sure doesn't sound like a guy who just found an ice wall that stretched endlessly around the entire flat earth, now does it?

Ah, I love selective quotes. Total Lackey is quoting from the accounts of Cook's second voyage. Please read the entire section. He was referring to what was called then "field ice" = floating pieces of ice of varying sizes, nowadays called pack ice. Cook went on to lament that the ice prevented him from sailing further south, and he speculated that beyond the pack ice lay a land which contained the notyetdiscovered South Pole. Keep in mind that from the deck of a ship all you can see is to about 8 miles away under perfect clear sky conditions. After that it's over the horizon. So he turned around and headed north. Thus is all from page 160 of the accounts of Cook's second voyage. Sure doesn't sound like a guy who just found an ice wall that stretched endlessly around the entire flat earth, now does it?
"It was ... an obstruction of such character as to leave no doubt in my mind as to our future proceedings, for we might as well sail through the cliffs of Dover as to penetrate such a mass.
It would be impossible to conceive a more solidlooking mass of ice; not the smallest appearance of any rent or fissure could we discover throughout its whole extent, and the intensely bright sky beyond it but too plainly indicated the great distance to which it reached southward."
—Sir James Clark Ross

Ross and Cook were sailing around in wooden boats in the 18th and 19th Centuries respectively. They didn't have access to skidoos, dogsleds C47s and De Havilland Twin Otters.
You might wish to consider further works of research (and television?) in the 20th and 21st Centuries. Could I suggest Pole to Pole (Michael Palin, BBC Books 1993) and the TV catalogue of Sir David Attenborough?

I have spent some time on boats also. They do use flat Maps. But they have a different map for the Southern Hemisphere and a different map for the Northern Hemisphere.
The mariners' operating in the Southern Hemisphere know their actual, physical travel distances as accurately as Mariners operating in the North.
I've never seen a flat Earth Map that accurately depicts travel distances in the southern hemisphere.

I forgot about an interesting “aroundtheworld” sailing race that’s been ongoing since 1989 occurring roughly every four years. It’s a grueling singlehanded nonstop race from a port in France (a French sailor is the founder of this race) called the Vendée Globe. They sail south to the Cape of Good Hope (southern tip of Africa), continue south to sail clockwise around Antarctica, then return to France, again nonstop, no outside assistance allowed. Typically takes about 9 months. There have been many boats from all different countries that have participated.
One of the major strategic decisions the skippers have to make is that if they sail closer to Antarctica this distance they have to travel is less since they’re basically circling Antarctica, so that saves time, but then the odds of running into ice is higher. And these guys have to sleep sometime. Given the speeds these pedigreed racing sailboats can travel now if they slam into a big enough chunk of ice  their boat is destroyed and they’re dead. They are not permitted any outside input like from meteorologists and satellite photos. It’s entirely up to each skipper. The typical sea conditions are 2030 foot waves and winds in the 4050 knot range. An amazing feat of endurance.
But how can this race exist if the typical FE model is true? And it happens about every four years with multiple boats. All the people involved are lying?

Also, these sailing races don't come out of nowhere: they take the "Clipper route" used by merchant sailing ships of the 19th century. Sailors of that time already knew they had to choose between a safer route or a shorter one further south, but with increased risks of meeting an iceberg.