What is the radius of the FE Earth model?

Math is indisputable and I feel this information would be really helpful for future FE discussion.

It doesn't have to be exact but a really close estimate would be great.

Tom Bishop you seem to have a wealth of knowledge regarding this theory. Any insight would be helpful. 

Please read the FAQ.

Also, please post questions in the Q&A forum.

Sorry about that. Thanks for the reply.

I checked it out and noticed that the values for diameter and circumference don't quite match up mathematically. I'm not sure which of the values to take as the correct.

Whenever there's a 'read the FAQ' response it means that the subject is discussed in the FAQ and alludes to some sort of answer, but really there is no answer, made up or not, there is no answer on flat earth stuff for anything.  The FAQ is merely a way to make you think they have some answers.

Think about this: the flat earth model is up in the air. Antarctica=continent or ice wall? They don't know this.  Obviously if they don't know this they're not gonna know the diameter of the earth.  Tom argues that in the FAQ it's talking about the known world.  No one's been to the ice wall or whatever and no one can be sure if they were at the ice wall or antarctica.  There's no way to prove the circumference is equal to what the diameter of the "known" world would suggest.

This is the problem with trying to make fantasy work logically.  Like the stupid Twilight books.  There are all these dead ends and no way to work through it logically.  There is so much fantasy in FET to try and really nail down any answers is futile.  Read the FAQ and you'll see the sun and moon are 32 miles across.  Come on... really?

Whenever there's a 'read the FAQ' response it means that the subject is discussed in the FAQ and alludes to some sort of answer, but really there is no answer, made up or not, there is no answer on flat earth stuff for anything.  The FAQ is merely a way to make you think they have some answers.

Think about this: the flat earth model is up in the air. Antarctica=continent or ice wall? They don't know this.  Obviously if they don't know this they're not gonna know the diameter of the earth.  Tom argues that in the FAQ it's talking about the known world.  No one's been to the ice wall or whatever and no one can be sure if they were at the ice wall or antarctica.  There's no way to prove the circumference is equal to what the diameter of the "known" world would suggest.

This is the problem with trying to make fantasy work logically.  Like the stupid Twilight books.  There are all these dead ends and no way to work through it logically.  There is so much fantasy in FET to try and really nail down any answers is futile.  Read the FAQ and you'll see the sun and moon are 32 miles across.  Come on... really?

I JUST raised this EXACT point in another forum.  I can not STAND it when someone starts a post, and a moderator or a moderator wannabe Nazi tells us to look at the FAQ or that the question has been raised in another forum.  It's like, excuse me... YOU'RE the ones with the fringe belief and the pseudo-science.  WE came to YOU for answers, and we don't feel like thumbing through pages and pages and pages of 2-year-old forum posts to find answers to the questions.  I've read the FAQ, and it sucks.  It's all circumstantial claims.  As a poster, I would be MUCH more apt to offering my opinion and insight to a new post with < 20 replies as opposed to a post that was made 3 years ago with 100 replies to thumb through.

I mean, it's like, this website is, for all intents and purposes, THE repository for Flat Earth education.  What if, every time someone asked someone else a scientific question, they either said, "Look it up" or "I already answered this question in a conversation 3 years ago."  What kind of elitist.... it's like, we're here because we wanna undersatnd.  I think FET is straight goofy, but I find it interesting.  Perhaps if you ANSWER this guy's question, someone could come back and refute it.  Or, you could just tell the guy to check the vague FAQ that doesn't really provide any kind of scientific answer so that nobody can refute what your reponse would have been.

And btw... how in the bloody freak can they not know the diameter of the earth, but they can say with certainty that the moon and sun are 32 miles across.

And btw... how in the bloody freak can they not know the diameter of the earth, but they can say with certainty that the moon and sun are 32 miles across.

We can see the sun and moon in their entirety.  Not so the Earth.

Not that I claim with certainty that the moon and sun are 32 miles across, but just to point out that there's no hole in the logic here.

I checked it out and noticed that the values for diameter and circumference don't quite match up mathematically. I'm not sure which of the values to take as the correct.

You assume the the flat Earth is a perfect circle.

Erasthonese's shadow experiment tells us the diameter of the known earth. I'm posting from a phone and can't link, but there's an article in the wiki under Form and Magnitude > Diameter on the subject.

Erasthonese's shadow experiment tells us the diameter of the known earth. I'm posting from a phone and can't link, but there's an article in the wiki under Form and Magnitude > Diameter on the subject.

It's his proof of a round Earth.
How can you use it to demonstrate a FE?

Quote from: Tom Bishop on August 20, 2012, 08:21:21 PM

Erasthonese's shadow experiment tells us the diameter of the known earth. I'm posting from a phone and can't link, but there's an article in the wiki under Form and Magnitude > Diameter on the subject.

It's his proof of a round Earth.
How can you use it to demonstrate a FE?

Much of chemistry was learned through the sisyphean efforts of alchemy. Should we throw away all that was learned because opposite results were achieved?

Quote from: EmperorZhark on August 21, 2012, 12:47:27 AM

It's his proof of a round Earth.
How can you use it to demonstrate a FE?

Much of chemistry was learned through the sisyphean efforts of alchemy. Should we throw away all that was learned because opposite results were achieved?

A.  A question was asked, which you apparently can't answer.

B.  Alchemists made observations on how different materials reacted, which were useful to know for chemistry.  Eratosthenes made observations about a round earth.  It could be used for FET, I suppose, but that brings us back to the question of how.

The wiki page is nonsensical.

I should also note that Eratosthenes' observations would place the sun approximately 4000 nautical miles above the flat earth.

I'd love to know how it can be used to find the circumference of the flat earth, though.

Quote from: KristaGurl on August 20, 2012, 01:32:57 PM

And btw... how in the bloody freak can they not know the diameter of the earth, but they can say with certainty that the moon and sun are 32 miles across.

We can see the sun and moon in their entirety.  Not so the Earth.

Not that I claim with certainty that the moon and sun are 32 miles across, but just to point out that there's no hole in the logic here.

If the sun and moon were round, you wouldn't be able to see them in their entirety, could you?

Besides... how do you determine the distance between the sun and moon?  Because... that is also incorrect.  When you use an incorrect measurement in an equation, the product of that equation is also incorrect. 

Erasthonese's shadow experiment tells us the diameter of the known earth.

Actually, the diameter of the earth that Eratosthenes calculated was far larger than what was know to the ancient Greeks.  I'm reasonably sure that they had no knowledge of the Americas, Australia and Antarctica or the full extent of Europe, Asia and Africa.

Quote from: Tom Bishop on August 20, 2012, 08:21:21 PM

Erasthonese's shadow experiment tells us the diameter of the known earth. I'm posting from a phone and can't link, but there's an article in the wiki under Form and Magnitude > Diameter on the subject.

It's his proof of a round Earth.
How can you use it to demonstrate a FE?

He's using shadows and sticks to estimate the surface of land the light of the sun affects.

I should also note that Eratosthenes' observations would place the sun approximately 4000 nautical miles above the flat earth.

I'd love to know how it can be used to find the circumference of the flat earth, though.

The difference is likely due to experimental error in Eratosthenes' time. The point was we can use his shadow experiment to estimate Flat Earth figures for the diameter of the earth and the height of the sun. Under his measurements the sun is 4000 miles above the earth.

Not gonna lie, trying to measure the diameter of Earth based on a somewhat small shadow sounds like a load of horse poop. I use the sun in many ways (sun lines, azimuth), but measuring the diameter of the whole Earth... far fetched if you ask me.

we can use his shadow experiment to estimate Flat Earth figures for the diameter of the earth

How?

Let's take a look at the wiki page here:

Syene and Alexandria are two North-South points with a distance of 500 nautical miles. Eratosthenes discovered through the shadow experiment that while the sun was exactly overhead of one city, it was 7°12' south of zenith at the other city.

7°12' makes a sweep of 1/25th of the FE's total longitude from 90°N to 90°S (radius).

Therefore we can take the distance of 500 nautical miles, multiply by 25, and find that the radius of the Flat Earth is about 12,250 nautical miles. Doubling that figure for the diameter we get a figure of 25,000 miles.

There are many problems here, but let's start with the bolded section first.  Obviously longitude should be latitude there.  It's a small mistake and who cares.

What I would like to know is how are lines of latitude related to the position of the sun in FET?

This article seems to be saying that if I go up 7 degrees, 12 minutes in latitude, that the sun's position will shift by that much.  If that's so, are lines of latitude evenly spaced apart in FET?

Secondly, how do you know these lines go to 90 degrees north and 90 degrees south?

Quote from: Roundy the Truthinessist on August 20, 2012, 04:37:23 PM

Quote from: KristaGurl on August 20, 2012, 01:32:57 PM

And btw... how in the bloody freak can they not know the diameter of the earth, but they can say with certainty that the moon and sun are 32 miles across.

We can see the sun and moon in their entirety.  Not so the Earth.

Not that I claim with certainty that the moon and sun are 32 miles across, but just to point out that there's no hole in the logic here.

If the sun and moon were round, you wouldn't be able to see them in their entirety, could you?

Don't be pedantic.  You'd be able to see enough to judge their dimensions.  I obviously don't need to be able to see all parts of a tennis ball at once to be able to measure that its diameter is about 2.63 inches.

Besides... how do you determine the distance between the sun and moon?  Because... that is also incorrect.  When you use an incorrect measurement in an equation, the product of that equation is also incorrect.

That is irrelevant to your original query.  You didn't ask how FEers can know the sizes of the sun and moon (and I agree, FE attempts to define them thus far have been wanting), you asked how FEers can claim to know the sizes of the sun and moon without knowing the size of the Earth.  That is what I was answering.

Don't be pedantic.  You'd be able to see enough to judge their dimensions.  I obviously don't need to be able to see all parts of a tennis ball at once to be able to measure that its diameter is about 2.63 inches.

Except that your estimates of the diameters of the sun/moon rely on knowing the distance to them, and oddly enough the answer to that question always comes up differently depending on who's doing the calculating and where they're doing it.

Quote from: Tom Bishop on August 21, 2012, 03:17:21 PM

we can use his shadow experiment to estimate Flat Earth figures for the diameter of the earth

How?

Let's take a look at the wiki page here:

Quote

Syene and Alexandria are two North-South points with a distance of 500 nautical miles. Eratosthenes discovered through the shadow experiment that while the sun was exactly overhead of one city, it was 7°12' south of zenith at the other city.

7°12' makes a sweep of 1/25th of the FE's total longitude from 90°N to 90°S (radius).

Therefore we can take the distance of 500 nautical miles, multiply by 25, and find that the radius of the Flat Earth is about 12,250 nautical miles. Doubling that figure for the diameter we get a figure of 25,000 miles.

There are many problems here, but let's start with the bolded section first.  Obviously longitude should be latitude there.  It's a small mistake and who cares.

What I would like to know is how are lines of latitude related to the position of the sun in FET?

This article seems to be saying that if I go up 7 degrees, 12 minutes in latitude, that the sun's position will shift by that much.  If that's so, are lines of latitude evenly spaced apart in FET?

Secondly, how do you know these lines go to 90 degrees north and 90 degrees south?

The lines of longitude are related to the position of the sun at noon equinox. At 0 degrees (equator) the sun is directly overhead. At 45 degrees North the sun has descended 45 degrees in the sky.

Quote from: Roundy the Truthinessist on August 21, 2012, 08:16:46 PM

Don't be pedantic.  You'd be able to see enough to judge their dimensions.  I obviously don't need to be able to see all parts of a tennis ball at once to be able to measure that its diameter is about 2.63 inches.

Except that your estimates of the diameters of the sun/moon rely on knowing the distance to them, and oddly enough the answer to that question always comes up differently depending on who's doing the calculating and where they're doing it.

On the earth's distance from the sun Copernicus computed it as 3,391,200 miles, Kepler contradicted him with an estimate of 12,376,800 miles, while Newton had asserted that it did not matter whether it was 28 million or 54 million miles 'for either will do as well'.

Quote from: Cat Earth Theory on August 21, 2012, 08:31:09 PM

Quote from: Roundy the Truthinessist on August 21, 2012, 08:16:46 PM

Don't be pedantic.  You'd be able to see enough to judge their dimensions.  I obviously don't need to be able to see all parts of a tennis ball at once to be able to measure that its diameter is about 2.63 inches.

Except that your estimates of the diameters of the sun/moon rely on knowing the distance to them, and oddly enough the answer to that question always comes up differently depending on who's doing the calculating and where they're doing it.

On the earth's distance from the sun Copernicus computed it as 3,391,200 miles, Kepler contradicted him with an estimate of 12,376,800 miles, while Newton had asserted that it did not matter whether it was 28 million or 54 million miles 'for either will do as well'.

newton said some crazy stuff.

By the way, what is the context for newton's assertion?

The lines of longitude are related to the position of the sun at noon equinox. At 0 degrees (equator) the sun is directly overhead. At 45 degrees North the sun has descended 45 degrees in the sky.

If longitude lines are related to the position of the sun at noon equinox, why are you talking about the equator and 45 degrees north?  Those are latitude lines.

Are the lines of latitude evenly spaced?  Because if they represent the position of the sun at noon on the equinox they can't be unless you want to throw out trigonometry (which you've done in the past, so it wouldn't surprise me).

Quote from: Cat Earth Theory on August 21, 2012, 04:50:28 PM

Quote from: Tom Bishop on August 21, 2012, 03:17:21 PM

we can use his shadow experiment to estimate Flat Earth figures for the diameter of the earth

How?

Let's take a look at the wiki page here:

Quote

Syene and Alexandria are two North-South points with a distance of 500 nautical miles. Eratosthenes discovered through the shadow experiment that while the sun was exactly overhead of one city, it was 7°12' south of zenith at the other city.

7°12' makes a sweep of 1/25th of the FE's total longitude from 90°N to 90°S (radius).

Therefore we can take the distance of 500 nautical miles, multiply by 25, and find that the radius of the Flat Earth is about 12,250 nautical miles. Doubling that figure for the diameter we get a figure of 25,000 miles.

There are many problems here, but let's start with the bolded section first.  Obviously longitude should be latitude there.  It's a small mistake and who cares.

What I would like to know is how are lines of latitude related to the position of the sun in FET?

This article seems to be saying that if I go up 7 degrees, 12 minutes in latitude, that the sun's position will shift by that much.  If that's so, are lines of latitude evenly spaced apart in FET?

Secondly, how do you know these lines go to 90 degrees north and 90 degrees south?

The lines of longitude are related to the position of the sun at noon equinox. At 0 degrees (equator) the sun is directly overhead. At 45 degrees North the sun has descended 45 degrees in the sky.

Tom that is sooooo wrong. The lines you are referring to are called latitude lines. You're using "rubbish arguments".

Quote from: EmperorZhark on August 21, 2012, 12:47:27 AM

Quote from: Tom Bishop on August 20, 2012, 08:21:21 PM

Erasthonese's shadow experiment tells us the diameter of the known earth. I'm posting from a phone and can't link, but there's an article in the wiki under Form and Magnitude > Diameter on the subject.

It's his proof of a round Earth.
How can you use it to demonstrate a FE?

He's using shadows and sticks to estimate the surface of land the light of the sun affects.

Let's see if I have this right.  On the days of the equinox the sun illuminates an area from 90 degrees north to 90 degrees south, correct?  If so, then it seems that on the days of the southern solstice that the sun should illuminate an area from about 66.5 degrees north to about 113.5 degrees south.  So, why does the "known" FE stop at 90 degrees south?

Math is indisputable and I feel this information would be really helpful for future FE discussion.

It doesn't have to be exact but a really close estimate would be great.

Tom Bishop you seem to have a wealth of knowledge regarding this theory. Any insight would be helpful.

I can give you a very precise estimate: R = 6356,66 kilometers

Here you will find out why (you will need some translation):

http://cercetare.forumgratuit.ro/t59p60-here-comes-the-sun#9224

It has everything to do with the rectification factor of the Great Pyramid, 286.1 sacred inches (7.272 meters)...


EDIT: I wrote initially meters instead of km...luckily I have a chance now to correct the mistake...

Quote from: Tom Bishop on August 22, 2012, 07:04:18 AM

The lines of longitude are related to the position of the sun at noon equinox. At 0 degrees (equator) the sun is directly overhead. At 45 degrees North the sun has descended 45 degrees in the sky.

If longitude lines are related to the position of the sun at noon equinox, why are you talking about the equator and 45 degrees north?  Those are latitude lines.

Are the lines of latitude evenly spaced?  Because if they represent the position of the sun at noon on the equinox they can't be unless you want to throw out trigonometry (which you've done in the past, so it wouldn't surprise me).

My mistake. You're correct. They're latitude lines.

Quote from: Tom Bishop on August 21, 2012, 03:17:21 PM

Quote from: EmperorZhark on August 21, 2012, 12:47:27 AM

Quote from: Tom Bishop on August 20, 2012, 08:21:21 PM

Erasthonese's shadow experiment tells us the diameter of the known earth. I'm posting from a phone and can't link, but there's an article in the wiki under Form and Magnitude > Diameter on the subject.

It's his proof of a round Earth.
How can you use it to demonstrate a FE?

He's using shadows and sticks to estimate the surface of land the light of the sun affects.

Let's see if I have this right.  On the days of the equinox the sun illuminates an area from 90 degrees north to 90 degrees south, correct?  If so, then it seems that on the days of the southern solstice that the sun should illuminate an area from about 66.5 degrees north to about 113.5 degrees south.  So, why does the "known" FE stop at 90 degrees south?

There are no degrees beyond 90^o N or 90^o S. The sun can only be a maximum angle of 90 degrees past zenith in the sky.

There are no degrees beyond 90^o N or 90^o S. The sun can only be a maximum angle of 90 degrees past zenith in the sky.

On the day of the southern solstice, the sun wouldn't reach 90 degrees past its zenith until an observer reaches about 113.5 degrees south.  113.5 degrees south sounds like a perfectly reasonable place to call "the edge of the known FE" rather than 90 degrees south.  After all, that is the illuminated region that Eratosthenes calculated, isn't it?

Quote from: Tom Bishop on August 29, 2012, 09:22:03 PM

There are no degrees beyond 90^o N or 90^o S. The sun can only be a maximum angle of 90 degrees past zenith in the sky.

On the day of the southern solstice, the sun wouldn't reach 90 degrees past its zenith until an observer reaches about 113.5 degrees south.  113.5 degrees south sounds like a perfectly reasonable place to call "the edge of the known FE" rather than 90 degrees south.  After all, that is the illuminated region that Eratosthenes calculated, isn't it?

The latitude lines are based on the sun's position on the day of the equinox when the sun is directly over the equator, not the day of the southern solstice.

Quote from: markjo on August 29, 2012, 09:51:08 PM

Quote from: Tom Bishop on August 29, 2012, 09:22:03 PM

There are no degrees beyond 90^o N or 90^o S. The sun can only be a maximum angle of 90 degrees past zenith in the sky.

On the day of the southern solstice, the sun wouldn't reach 90 degrees past its zenith until an observer reaches about 113.5 degrees south.  113.5 degrees south sounds like a perfectly reasonable place to call "the edge of the known FE" rather than 90 degrees south.  After all, that is the illuminated region that Eratosthenes calculated, isn't it?

The latitude lines are based on the sun's position on the day of the equinox when the sun is directly over the equator, not the day of the southern solstice.

Irrelevant.  The sun is not the only celestial body that can be used to determine latitude.  In fact, one should be able to use the southern constellations to reliably navigate significantly further south than 113.5 degrees.