Distances on RE and FE consistent thanks to bendy light.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #300 on: February 08, 2011, 02:23:21 PM »
I think what he means is that the notion that a flat earth would have the same distances that we can readily measure and see in life would show different results that what we have obtained through observation.  For instance, how can I sail around the world through the antarctic and the arctic at mirrored latitudes and still have the same travel time?
Yes, essentially.
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silver

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #301 on: February 08, 2011, 02:54:15 PM »
I think what he means is that the notion that a flat earth would have the same distances that we can readily measure and see in life would show different results that what we have obtained through observation.  For instance, how can I sail around the world through the antarctic and the arctic at mirrored latitudes and still have the same travel time?
Yes, essentially.

So you're admitting that the OP is wrong? :o

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #302 on: February 08, 2011, 02:58:19 PM »
So you're admitting that the OP is wrong? :o
No, why?
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #303 on: February 08, 2011, 05:54:47 PM »
I missed the part when PP realized that no maps have to be considered for a counter arguement.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #304 on: February 08, 2011, 06:19:59 PM »
I missed the part when PP realized that no maps have to be considered for a counter arguement.
I know you did, Mr "Caltech". You have yet to make such an argument, with all the specifications I have mentioned in mind.
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29silhouette

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #305 on: February 08, 2011, 08:10:56 PM »

Why should FE map need to account for bendy light?  The true coordinates of locations in flat earth would easily be represented on a flat map.
Only in 1:1 scale. Any other scale would have to be re-adjusted to scale down light's bending, thus resulting in a hollow object of a sort. Any other scale mapped on a flat surface is going to be imprecise, and the precision is going to be inversely proportional to how far from 1:1 we go.
If an accurate map were made of a flat earth using aerial photography from a height low enough that bendy light (if it existed) had no influence on the pictures, and all the pictures were put together into one big image, this one image could be enlarged or shrunk, and it would still accurately show the layout of landmasses whether it's 1:1 or 1:150,000.  And as far as I can tell, there are aerial photos of the entire planet (or most of it), so let's start making this FE map.


Madagascar and Australia are 6500 km apart.  How can this be true for Flat Earth?
For example, they could be 6500km apart. That's how.
Apples exist on the Earth. How can this be true for the Earth?
It's also about 6500km from the Bering Straight to the southern tip of Texas.  Does the FE map show both of those distances the same? 

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forreal

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #306 on: February 08, 2011, 08:24:36 PM »

Why should FE map need to account for bendy light?  The true coordinates of locations in flat earth would easily be represented on a flat map.
Only in 1:1 scale. Any other scale would have to be re-adjusted to scale down light's bending, thus resulting in a hollow object of a sort. Any other scale mapped on a flat surface is going to be imprecise, and the precision is going to be inversely proportional to how far from 1:1 we go.
If an accurate map were made of a flat earth using aerial photography from a height low enough that bendy light (if it existed) had no influence on the pictures, and all the pictures were put together into one big image, this one image could be enlarged or shrunk, and it would still accurately show the layout of landmasses whether it's 1:1 or 1:150,000.  And as far as I can tell, there are aerial photos of the entire planet (or most of it), so let's start making this FE map.

Since we do have a large accumulation of pictures of the earths surface, how can it be that we don't have an abundance of photographs for the southern regions of the world?  In a FE world there would be a much larger spread of imagery to compensate for the larger distance to complete an aerial revolution around the world.

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Around And About

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #307 on: February 09, 2011, 12:28:32 AM »
PizzaPlanet, I appreciate your timely response (delays are fine, I don't always have immediate access, myself), but I honestly don't understand it. Disregarding your statements concerning the nature of light, since my theoretical models have no light sources, all I could really discern was that the distances were somehow essentially the same for both models. If that is true, it seems logical to conclude that:



...where N = North Pole, S = South Pole, and C = Center. The darker lines represent greater circumferences and the lighter shades represent lesser circumferences. It seems to hold true for RET, is my representation of FET accurate?

Essentially, my question is....past the equatorial line on FET's model, how can you proceed outward away from the center and achieve smaller circumferences?
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #308 on: February 09, 2011, 04:02:12 AM »
I missed the part when PP realized that no maps have to be considered for a counter arguement.
I know you did, Mr "Caltech". You have yet to make such an argument, with all the specifications I have mentioned in mind.

I told you one.
your counter-counter point was that I used maps. you have yet to show how one would use the FET model to approximate the area of colorado.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #309 on: February 09, 2011, 04:13:42 AM »
you have yet to show how one would use the FET model to approximate the area of colorado.
Wow. I would have thought a person that spent a half of their FES time repeating the word "odometer" over and over would know what an odometer is.
Also, sup with the Colorado obsession?

PizzaPlanet, I appreciate your timely response (delays are fine, I don't always have immediate access, myself), but I honestly don't understand it. Disregarding your statements concerning the nature of light, since my theoretical models have no light sources, all I could really discern was that the distances were somehow essentially the same for both models. If that is true, it seems logical to conclude that:



...where N = North Pole, S = South Pole, and C = Center. The darker lines represent greater circumferences and the lighter shades represent lesser circumferences. It seems to hold true for RET, is my representation of FET accurate?

Essentially, my question is....past the equatorial line on FET's model, how can you proceed outward away from the center and achieve smaller circumferences?
Being very concise, my idea assumes a non-Euclidean geometric model. Of course, it would all be completely impossible, were it not for that.
« Last Edit: February 09, 2011, 04:15:45 AM by PizzaPlanet »
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Nolhekh

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #310 on: February 09, 2011, 06:10:02 AM »
Being very concise, my idea assumes a non-Euclidean geometric model. Of course, it would all be completely impossible, were it not for that.

If you're going to use a non-euclidean model, please specify and define the model and show how it applies to reality, before basing your arguments on it.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #311 on: February 09, 2011, 06:23:56 AM »
If you're going to use a non-euclidean model, please specify and define the model and show how it applies to reality, before basing your arguments on it.
OP
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Nolhekh

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #312 on: February 09, 2011, 06:49:26 AM »
If you're going to use a non-euclidean model, please specify and define the model and show how it applies to reality, before basing your arguments on it.
OP

I see a diagram showing light bending then hitting the earth at different angles.  The angle at which light hits the ground does not affect the appearance of distances.  It is the light reflecting off the ground and into your eyes that would affect the appearance of distance if it was bent.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #313 on: February 09, 2011, 07:35:53 AM »
The angle at which light hits the ground does not affect the appearance of distances.
Incorrect.

It is the light reflecting off the ground and into your eyes that would affect the appearance of distance if it was bent.
Which is directly affected by the former.
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forreal

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #314 on: February 09, 2011, 11:17:30 AM »
The angle at which light hits the ground does not affect the appearance of distances.
Incorrect.

It is the light reflecting off the ground and into your eyes that would affect the appearance of distance if it was bent.
Which is directly affected by the former.

But that doesn't affect the distance required to travel a set pathway.  Even if we assumed the FET held true, the southern hemispheres appearance due to bendy light has no effect whatsoever on its actual size.  So how you say that the world is HUGE around the southern hemisphere, when there are people like me that have sailed around it, and can honestly say that it is not as large as the FET model would require it to be?

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Beorn

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #315 on: February 09, 2011, 11:32:02 AM »
I have sailed around it and it was exactly as big as the FET model would require it to be.
Quote
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Hessy

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #316 on: February 09, 2011, 11:36:30 AM »
I have sailed around it and it was exactly as big as the FET model would require it to be.

Do you have any evidence to support this outlandish claim?

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #317 on: February 09, 2011, 11:48:34 AM »
But that doesn't affect the distance required to travel a set pathway.  Even if we assumed the FET held true, the southern hemispheres appearance due to bendy light has no effect whatsoever on its actual size.
Of course. Hence the thread, and its validity.

So how you say that the world is HUGE around the southern hemisphere
Which I do not

when there are people like me that have sailed around it
Ah, another sailor. Put him next to the astronomers.

and can honestly say that it is not as large as the FET model would require it to be?
Your honesty is very doubtful, yet irrelevant to the thread for reason stated above.
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forreal

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #318 on: February 09, 2011, 11:52:58 AM »
But that doesn't affect the distance required to travel a set pathway.  Even if we assumed the FET held true, the southern hemispheres appearance due to bendy light has no effect whatsoever on its actual size.
Of course. Hence the thread, and its validity.

So how you say that the world is HUGE around the southern hemisphere
Which I do not

when there are people like me that have sailed around it
Ah, another sailor. Put him next to the astronomers.

and can honestly say that it is not as large as the FET model would require it to be?
Your honesty is very doubtful, yet irrelevant to the thread for reason stated above.

None of your answers came close to answering any of the questions which I brought up.  I'll make it simple this time:

FET maps have enormous distances for circumnavigation of the southern regions of the planet.  How can that hold true while so many people have experienced travelling around the southern hemisphere and not taken the amount of time that would be required to traverse the FET model planet?

And again in even simper terms:

I think your map is broked.

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Nolhekh

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #319 on: February 09, 2011, 12:07:56 PM »
The angle at which light hits the ground does not affect the appearance of distances.
Incorrect.
Explain.

If I shine a flashlight at the floor at an angle, the section of the floor does not appear longer than it really is.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #320 on: February 09, 2011, 01:34:32 PM »
FET maps
There's your problem. The whole thread is about how maps do not prove anything. I hope these terms are simple enough for you.

If I shine a flashlight at the floor at an angle, the section of the floor does not appear longer than it really is.
Of course. That's because the path your light has taken from the source to its destination was, frankly, not very long. This is also exactly why maps would have to account for bendy light in any other scale than 1:1. Thanks for supporting my point with a personally conducted experiment.
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Nolhekh

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #321 on: February 09, 2011, 01:51:51 PM »
If I shine a flashlight at the floor at an angle, the section of the floor does not appear longer than it really is.
Of course. That's because the path your light has taken from the source to its destination was, frankly, not very long. This is also exactly why maps would have to account for bendy light in any other scale than 1:1. Thanks for supporting my point with a personally conducted experiment.

You imply here:
The angle at which light hits the ground does not affect the appearance of distances.
Incorrect.
that the angle at which the sunlight hits the ground affects the appearance of distances.
Now you're implying the distance the light travelled somehow changes the appearance of distances.  If sunlight strikes the ground 1 meter away from me, how can the angle at which it does this affect how far away from me it appears to strike?

Although I'm starting to see what you're getting at now... I think.  Let me try to simplify it.

If you make a map of the flat earth at 1:1, bendy light (however it works) will make the map itself appear the same way the earth does, but if the map is shrunk, bendy light will not make it look exactly like the real thing, as it will look distorted.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #322 on: February 09, 2011, 03:27:56 PM »
If you make a map of the flat earth at 1:1, bendy light (however it works) will make the map itself appear the same way the earth does, but if the map is shrunk, bendy light will not make it look exactly like the real thing, as it will look distorted.
Yes, that's exactly it
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forreal

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #323 on: February 09, 2011, 06:11:51 PM »
Quote
There's your problem. The whole thread is about how maps do not prove anything. I hope these terms are simple enough for you.[\quote]

The below is not a quote - darn forums code is buggy.

I think you are slow.  I am saying how can this planet be flat if it takes me the same time/distance to circumnavigate the globe in the southern and northern hemispheres?  In the FET model they would be separate distances from the center point (ie north pole) and thus have different circumferences.  Do you have any explanation for this?  And don't say its f***ing light.
« Last Edit: February 09, 2011, 06:13:50 PM by forreal »

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Around And About

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #324 on: February 09, 2011, 06:47:45 PM »
you have yet to show how one would use the FET model to approximate the area of colorado.
Wow. I would have thought a person that spent a half of their FES time repeating the word "odometer" over and over would know what an odometer is.
Also, sup with the Colorado obsession?

PizzaPlanet, I appreciate your timely response (delays are fine, I don't always have immediate access, myself), but I honestly don't understand it. Disregarding your statements concerning the nature of light, since my theoretical models have no light sources, all I could really discern was that the distances were somehow essentially the same for both models. If that is true, it seems logical to conclude that:



...where N = North Pole, S = South Pole, and C = Center. The darker lines represent greater circumferences and the lighter shades represent lesser circumferences. It seems to hold true for RET, is my representation of FET accurate?

Essentially, my question is....past the equatorial line on FET's model, how can you proceed outward away from the center and achieve smaller circumferences?
Being very concise, my idea assumes a non-Euclidean geometric model. Of course, it would all be completely impossible, were it not for that.


Hmm okay...that's perfectly fine, and I'm not really familiar with non-Euclidean geometry anyway. I just wish there was a way to actually picture the earth's structure proposed under FET in my head, since this in itself would probably clear up some other questions.
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29silhouette

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #325 on: February 09, 2011, 08:01:54 PM »
If you make a map of the flat earth at 1:1, bendy light (however it works) will make the map itself appear the same way the earth does, but if the map is shrunk, bendy light will not make it look exactly like the real thing, as it will look distorted.
Yes, that's exactly it
Bendy maps...interesting.

So if you have an image on paper that is 1:1 scale, and digitally or somehow physically shrink it (while maintaining the aspect ratio) down to 1:100, 1:1000, or 1:100,000, etc, bendy light will actually re-arrange the shapes of everything on it.  Do I have that right?

Might have to take all those aerial photos of the majority of the planet, and lay them out flat or compile them in some kind of image software to make that flat earth image so we can all see what it looks like.  Bendy light should have no effect on that at all.

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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #326 on: February 10, 2011, 12:12:13 AM »
i got too lazy to read the earlier posts, 17 pages is a lot

but, pizza, i want you to draw 3 parrellel lines on a orange peel the goes all the way around and meets itself again (aka, draw a circle, that are parellel.) one in the middle, and one equidistant from the top and the center, one equidistant from the bottom to the center
now i want you to mesure the length around the circles, do this to your best estimate

you should find the equator is longer than the ones near the pole

then cut it perpendicular to the lines with only one cut, that expands half of the whole circumfrence, in which it starts and ends at the poles

now, i want you to make that rectangular, without additional cuts

if you succeed, you will have 3 lines, that are of equal lenghts, but becuase the outer lines represent a smaller distance, due to a smaller original circumfrence, you must compensate for this

hence, if you were to divide the rectangle into fourths, a fourth of the equator, would not be the same length across as a fourth of the outer lines, hence your 1000 km to 200 km

google is makeing the round earth rectangular, but now, it must compensate for the increased length at the poles

hope you can understand this
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #327 on: February 10, 2011, 03:07:19 AM »
you have yet to show how one would use the FET model to approximate the area of colorado.
Wow. I would have thought a person that spent a half of their FES time repeating the word "odometer" over and over would know what an odometer is.
Also, sup with the Colorado obsession?


so I'm guessing you can't. I'm not telling you to measure it, but to approximate its area using the description of the flat earth. from the faq. its really simple. Just double integrate. And I'm using Colorado, because it was designed to be a perfect angular sector

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #328 on: February 10, 2011, 05:55:57 AM »
The below is not a quote - darn forums code is buggy.
It's not. Let's have a look at what you've done here:

Code: [Select]
[quote]There's your problem. The [u]whole thread[/u] is about how maps [b]do not[/b] prove anything. I hope these terms are simple enough for you.[\quote]Now, if I may bring your attention to the [\quote]. What you want is
Code: [Select]
[/quote]
I think you might be quite slow.

I think you are slow.
Oh my, oh my. Our feelings are mutual! Needless to say, your ad hominems have no place here.

I am saying how can this planet be flat if it takes me the same time/distance to circumnavigate the globe in the southern and northern hemispheres?
You're thinking in Euclidean geometries, which proves you haven't read the thread. Read the thread.

In the FET model they would be separate distances from the center point (ie north pole) and thus have different circumferences.
No.

Do you have any explanation for this?
Yes. It doesn't happen.

And don't say its f***ing light.
Ah, and here's my evidence that you are, indeed, slow. Light makes distances appear different, whilst they're not.

I'm not telling you to measure it, but to approximate its area using the description of the flat earth. from the faq.
So, a description I don't support? No, thanks.

its really simple. Just double integrate.
Integration? In this geometry? Really?

And I'm using Colorado, because it was designed to be a perfect angular sector
You have yet to point me to a valid disproof. Right now you are not using Colorado for anything of significance, importance, nor relevancy to the thread.
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #329 on: February 10, 2011, 06:30:11 AM »
The below is not a quote - darn forums code is buggy.
It's not. Let's have a look at what you've done here:

Code: [Select]
[quote]There's your problem. The [u]whole thread[/u] is about how maps [b]do not[/b] prove anything. I hope these terms are simple enough for you.[\quote]Now, if I may bring your attention to the [\quote]. What you want is
Code: [Select]
[/quote]
I think you might be quite slow.

I think you are slow.
Oh my, oh my. Our feelings are mutual! Needless to say, your ad hominems have no place here.

I am saying how can this planet be flat if it takes me the same time/distance to circumnavigate the globe in the southern and northern hemispheres?
You're thinking in Euclidean geometries, which proves you haven't read the thread. Read the thread.

In the FET model they would be separate distances from the center point (ie north pole) and thus have different circumferences.
No.

Do you have any explanation for this?
Yes. It doesn't happen.

And don't say its f***ing light.
Ah, and here's my evidence that you are, indeed, slow. Light makes distances appear different, whilst they're not.

I'm not telling you to measure it, but to approximate its area using the description of the flat earth. from the faq.
So, a description I don't support? No, thanks.

its really simple. Just double integrate.
Integration? In this geometry? Really?

And I'm using Colorado, because it was designed to be a perfect angular sector
You have yet to point me to a valid disproof. Right now you are not using Colorado for anything of significance, importance, nor relevancy to the thread.

Now you're just being intentionally slow.

I said use the description from the FAQ and wikipedia. its a physical description. I'll tell  you why its relevant if you can't see why. Use the Lat. points as the initial and final radii, and use long. for the initial and final angles. double integrate that shit with respect to radius and angle. What do you get?

And I use doubs integral, because its convenient for wolfram alpha, as opposed to using proportions and formulas