Author Topic: Flat Earth-curious (Read 12483 times)

Ladislaus · « **Reply #135 on:** January 10, 2022, 05:49:41 AM »

Quote from: andy on January 08, 2022, 11:15:45 PM

High energy theories indeed fail to describe the universe but in the world "near us" all calculations are so damn precise. This is not a secret.

Is NASA hiding stuff? You bet. But they do not lie about everything.

Yes, the math works, but the math merely describes things. Math can describe the rate at which something falls, but the actual cause for why it falls has remained elusve. Gravity has never been proven, and there's a wide array of competing theories about what causes the phenomenon. Tesla rightly pointed out that a fantasy world has been built up with one equation layered on top of another.

NASA lies about a LOT of stuff, and that's enough to discredit them as an authority. I don't have to prove that they lie about everything. But it isn't just one or two things; their entire space program is a complete fraud.

Ladislaus · « **Reply #136 on:** January 10, 2022, 05:59:12 AM »

If people want, math and physics, check out Dr. John D.
https://www.cathinfo.com/fighting-errors-in-the-modern-world/dr-john-d-destroys-globe-earth/

Very few responses. This man is less popular than the other youtubers because, quite frankly, his channel is extremely dry. And it's very dry because he does all the math and runs the numbers, so his laser test video goes over an hour, and your average viewer doesn't have the patience for it.

I noticed that there were zero glober refutations.

He destroys refraction by conducting a two-way laser test. He took all the temperature and humidity readings, and recorded all the numbers with precision. Really the only comeback that globers have against all the "see too far" experiments is the gratuitous mention of the magic word "refraction". It's never proven or demonstrated. In order for light to follow exactly the cuvature of the globe, there would have to be an increasing density gradient along the path of the laser, as Dr. John D points out. But if there's an increasing density gradient, then in the reverse direction, there would be a decreasing density gradient. Consequently, the laser could not work in both directions at the same time at those distances. This puts the nail in the coffin of the "refraction" magic.

He also affirms my #2 reason why the current cosmology is bogus. Gravity is simply not strong enough to explain why the immense vacuum of space, as modern science conceives of it, would not rip the atmosphere off the planet. It's simply not possible that our atmosphere could exist adjacent to a nearly-perfect vacuum.

Anne Evergreen · « **Reply #137 on:** January 10, 2022, 06:18:49 PM »

Quote from: Miser Peccator on January 10, 2022, 03:07:07 AM

At what distance?

How many miles out?

Never bothered to measure. 15-20 miles, maybe? It would depend on the height of the mast for one thing, and then you could get into trig to calculate certain things I suspect.

It is the same situation with driving across flat prairie land where you see the approach of grain elevators getting bigger and bigger as you get closer and closer to them.

Miser Peccator · « **Reply #138 on:** January 10, 2022, 06:28:44 PM »

Quote from: Anne Evergreen on January 10, 2022, 06:18:49 PM

Never bothered to measure. 15-20 miles, maybe? It would depend on the height of the mast for one thing, and then you could get into trig to calculate certain things I suspect.

It is the same situation with driving across flat prairie land where you see the approach of grain elevators getting bigger and bigger as you get closer and closer to them.

Thanks

Can you or anyone else here who prefer the Ball Earth theory find out how far out the average ship needs to get?

Say an average person is on the beach watching a ship go out to sea.

What does the Ball Earth theory say?

I'm sure after centuries of this and with all of the technology available it has been measured accurately right?

I mean on a clear day.

I just want to know when the actual physical obstruction the curve of earth kicks in.

Dankward · « **Reply #139 on:** January 10, 2022, 06:47:03 PM »

Quote from: Meg on January 09, 2022, 10:37:12 AM

It might seem to have to be insane, yes, I understand that. But not impossible. We don't really know how it works exactly. And it will be difficult to provide you with an absolute working model for how it works, because it doesn't exist as yet. And it may not ever exist, unless NASA 'fesses up and tells the truth.

Sorry, I didn't watch the video in the link. Unless it addresses the specific issue of how the sun cannot work on a flat earth (and provide sunset light for both the north and south pole), I'd rather not go to the trouble of viewing it. I already know that there is a sunset sun 24 hours a day in Antarctica. I don't dispute that.

Good. We know really well how it works exactly, we understand the seasons and sunlight very well. But the good thing is that even if "NASA" (most science) was totally off, the scientific method would allow us to gradually gather new evidence until we have figured out a complete, working model.
"In an ideal world, flat Earthers would some day once again figure out that earth is in fact a spheroid" - unless they keep holding themselves back by biases.

This is what 24h Sun looks like on flat earth, this is the winter Sun lighting FE:

FE vs. GE

What do you think, which model explains it better?

Because the FE animation is what would have to happen on a flat surface, regardless of the map being a EAP map (globe projection).

Dankward · « **Reply #140 on:** January 10, 2022, 07:18:29 PM »

Quote from: Miser Peccator on January 10, 2022, 03:07:07 AM

At what distance?

How many miles out?

What height is the ship, what height is the observer (eye height)? We need a few more parameters.

When we have those, we can calculate it as described here: https://physics.stackexchange.com/questions/593064/how-to-calculate-how-much-of-an-object-is-hidden-due-to-earth-curvature

Earth's radius is 3959mi or 6371km.

Or we use this simpler formula: h = 6,371km * (1 - cos(0.009° * d))
We could solve for d here if h is known (too tired for that right now).
Let's assume the observer is on water level, and the ship is 30m high.

6371 * (1 - cos(0,009 * 19,55)) = 30m
So that gives us roughly d = 19,55km fo ar distance with the assumption of no atmospheric effects going on for the ship to stay visible longer or shorter.

Miser Peccator · « **Reply #141 on:** January 10, 2022, 07:24:44 PM »

Quote from: Dankward on January 10, 2022, 07:18:29 PM

What height is the ship, what height is the observer (eye height)? We need a few more parameters.

When we have those, we can calculate it as described here: https://physics.stackexchange.com/questions/593064/how-to-calculate-how-much-of-an-object-is-hidden-due-to-earth-curvature

Earth's radius is 3959mi or 6371km.

Or we use this simpler formula: h = 6,371km * (1 - cos(0.009° * d))
We could solve for d here if h is known (too tired for that right now).
Let's assume the observer is on water level, and the ship is 30m high.

6371 * (1 - cos(0,009 * 19,55)) = 30m
So that gives us roughly d = 19,55km fo ar distance with the assumption of no atmospheric effects going on for the ship to stay visible longer or shorter.

Is there an earth curve calculator we can use?

Anne Evergreen · « **Reply #142 on:** January 10, 2022, 07:43:04 PM »

Quote from: Miser Peccator on January 10, 2022, 06:28:44 PM

Thanks

Can you or anyone else here who prefer the Ball Earth theory find out how far out the average ship needs to get?

Say an average person is on the beach watching a ship go out to sea.

What does the Ball Earth theory say?

I'm sure after centuries of this and with all of the technology available it has been measured accurately right?

I mean on a clear day.

I just want to know when the actual physical obstruction the curve of earth kicks in.

You're welcome, but it won't be me to provide you with that. I mentioned something briefly to my husband (former pilot when he was younger) and he is too busy to bother with any of this. He mentioned a few things about wind speed, knots, height of the mast and so on, and I tuned out.

Ask another physics and math geek.

Anne Evergreen · « **Reply #143 on:** January 10, 2022, 07:46:39 PM »

Quote from: Dankward on January 10, 2022, 06:47:03 PM

Good. We know really well how it works exactly, we understand the seasons and sunlight very well. But the good thing is that even if "NASA" (most science) was totally off, the scientific method would allow us to gradually gather new evidence until we have figured out a complete, working model.
"In an ideal world, flat Earthers would some day once again figure out that earth is in fact a spheroid" - unless they keep holding themselves back by biases.

This is what 24h Sun looks like on flat earth, this is the winter Sun lighting FE:

FE vs. GE

What do you think, which model explains it better?

Because the FE animation is what would have to happen on a flat surface, regardless of the map being a EAP map (globe projection).

No question the bottom one. That's all I need to know. Thanks for this animation!

Ladislaus · « **Reply #144 on:** January 10, 2022, 07:51:12 PM »

#1, there is no 24-hour sun in Antarctica.
#2, the climate of Antarctica is massively colder than that of its opposite "pole" which it should not be based on the tilt model
#3, that first one is not an accurate picture of FE, which holds that there's a firmament which reflects the sun when it hits its outer limts, which does cause there to be more sun in the Antarctic, but never a 24-hour sun.

There was a timelapse video made pretending to show the 24-hour sun from Antarctica but it was proven to be a hoax due to absolutely identical cloud detail at the beginning and at the end of the video. Why fake it if there's a real 24-hour sun.

Anne Evergreen · « **Reply #145 on:** January 10, 2022, 07:55:58 PM »

Quote from: Dankward on January 10, 2022, 07:18:29 PM

What height is the ship, what height is the observer (eye height)? We need a few more parameters.

When we have those, we can calculate it as described here: https://physics.stackexchange.com/questions/593064/how-to-calculate-how-much-of-an-object-is-hidden-due-to-earth-curvature

Earth's radius is 3959mi or 6371km.

Or we use this simpler formula: h = 6,371km * (1 - cos(0.009° * d))
We could solve for d here if h is known (too tired for that right now).
Let's assume the observer is on water level, and the ship is 30m high.

6371 * (1 - cos(0,009 * 19,55)) = 30m
So that gives us roughly d = 19,55km fo ar distance with the assumption of no atmospheric effects going on for the ship to stay visible longer or shorter.

Alright how about a woman that is 6' tall, standing level at 650' above sea level. There is 100% visibility, no clouds, completely blue sky, June 1, above the equator, with a cruise ship that is sailing to Alaska from the port of Los Angeles,
CA up to Juneau, AK
The cruise ship is a maximum of 217' in height, and is travelling at maximum speed of 22 knots.

Anything else you need, you can add in.
Okay just going to add in here that there are no buildings around, and the nearest tree is 15' away and is 150' tall. It is 4PM in the afternoon.

Too tired to think of anything else. Thanks.

Ladislaus · « **Reply #146 on:** January 10, 2022, 07:58:28 PM »

Here's another proof of FE, that reality is closer to the Azimuthal equidistant map than the fake globe model.

FEs hold that the reality looks something like this (Azimuthal equidistant map from the North pole).

Mostly reconizable, right? You can make everythig out. But globe earthers allege that the Southern hemisphere continents are too big.

But is that the case? No. That perception is actually based on the distorted Mercator map. To correct this, they released the Galls-Peter projection map, which asserts that this shows the TRUE SIZE of the continents in the Southern hemisphere, and that they were deliberately shrunk on the Mercator map. They have their true size, even if, they assert, their shapes are slightly off in order to get the right size projection.

So now the size of the continents on the North Pole Azimuthal Equidistant map are so far off, are they?

And, if the earth were a true globe, a South Pole Azimuthal Equidistant map would be recognizable as well, just like the North Pole one is.

Well, have a look:

This is riduclous, and apart from Australia, South America, and to a degree, Africa, the rest is absolutely unrecognizable. But if the earth were a true globe, both of them would be rather recognizable and would be off by about the same degree.

This demonstrates that the Azimuthal Equidistant map from the North Pole is actually an extremely accurate representation of the TRUE layout of the world.

Miser Peccator · « **Reply #147 on:** January 10, 2022, 08:03:47 PM »

Quote from: Dankward on January 10, 2022, 07:18:29 PM

What height is the ship, what height is the observer (eye height)? We need a few more parameters.

When we have those, we can calculate it as described here: https://physics.stackexchange.com/questions/593064/how-to-calculate-how-much-of-an-object-is-hidden-due-to-earth-curvature

Earth's radius is 3959mi or 6371km.

Or we use this simpler formula: h = 6,371km * (1 - cos(0.009° * d))
We could solve for d here if h is known (too tired for that right now).
Let's assume the observer is on water level, and the ship is 30m high.

6371 * (1 - cos(0,009 * 19,55)) = 30m
So that gives us roughly d = 19,55km fo ar distance with the assumption of no atmospheric effects going on for the ship to stay visible longer or shorter.

So the curve is steep enough to hide the ship at about 12 miles out in this case?

Miser Peccator · « **Reply #148 on:** January 10, 2022, 08:15:49 PM »

Quote from: Anne Evergreen on January 10, 2022, 07:55:58 PM

Alright how about a woman that is 6' tall, standing level at 650' above sea level. There is 100% visibility, no clouds, completely blue sky, June 1, above the equator, with a cruise ship that is sailing to Alaska from the port of Los Angeles,
CA up to Juneau, AK
The cruise ship is a maximum of 217' in height, and is travelling at maximum speed of 22 knots.

Anything else you need, you can add in.
Okay just going to add in here that there are no buildings around, and the nearest tree is 15' away and is 150' tall. It is 4PM in the afternoon.

Too tired to think of anything else. Thanks.

LOL

That's too complicated for me!

I just want a simple answer.

Average person on the beach watching average ship go over curve.

How far out?

Anne Evergreen · « **Reply #149 on:** January 10, 2022, 08:28:37 PM »

Quote from: Ladislaus on January 10, 2022, 07:51:12 PM

#1, there is no 24-hour sun in Antarctica.
#2, the climate of Antarctica is massively colder than that of its opposite "pole" which it should not be based on the tilt model
#3, that first one is not an accurate picture of FE, which holds that there's a firmament which reflects the sun when it hits its outer limts, which does cause there to be more sun in the Antarctic, but never a 24-hour sun.

There was a timelapse video made pretending to show the 24-hour sun from Antarctica but it was proven to be a hoax due to absolutely identical cloud detail at the beginning and at the end of the video. Why fake it if there's a real 24-hour sun.

Brain is already mush for today, back is bugging me from shoveling, and I could only wish there were more sun around here! I don't trust a whole lot of what NASA says, but it cannot all be garbage. A lot of what was at Kennedy Space Centre made sense, and a lot did not. But there were an awful lot of "burnt" props, if you will, from something very hot, (ie. the sun), so however high the astronauts can go, I really do not know! Sounding like Dr. Seuss here.

There are all sorts of reasons for people to fake a whole lot of things, and money immediately comes to mind. But if you actually visit the Kennedy Space Centre, it seems there is an awful lot of money being spent to simply try and fool everyone.

So while I appreciate your thoughts and help, I personally am going to stick with the globe theory, and again, I have given many reasons why. Not the least of which are the pictures showing Jesus Christ King with a globe, or orb in His hand. If those pictures are wrong, then giving them veneration is blasphemy.

Since I am going to go with the assumption that you would not partake in that, HOW do you explain the pictures of Our Lord with a globe? You can't.

If the earth were flat, He would be shown with a flat map in His hands, right? So either one has to deny the Kingship of Christ (atheist), or accept that those pictures are acceptable for Catholics.

Therefore if anything, I would say the Flat Earth belief is for atheists. It does not speak to my natural instincts for understanding the world.

Author Topic: Flat Earth-curious (Read 12483 times)

Ladislaus

Re: Flat Earth-curious

Ladislaus

Re: Flat Earth-curious

Anne Evergreen

Re: Flat Earth-curious

Miser Peccator

Re: Flat Earth-curious

Dankward

Re: Flat Earth-curious

Dankward

Re: Flat Earth-curious

Miser Peccator

Re: Flat Earth-curious

Anne Evergreen

Re: Flat Earth-curious

Anne Evergreen

Re: Flat Earth-curious

Ladislaus

Re: Flat Earth-curious

Anne Evergreen

Re: Flat Earth-curious

Ladislaus

Re: Flat Earth-curious

Miser Peccator

Re: Flat Earth-curious

Miser Peccator

Re: Flat Earth-curious

Anne Evergreen

Re: Flat Earth-curious