Distances on RE and FE consistent thanks to bendy light.

Quote from: silver on February 16, 2011, 12:10:03 PM

The physical circumference of the south and north pole is the same. That is why it's relevant.

The geometry is not Euclidean. That's why it isn't.

Wait a minute.



The rim and the outer north pole have the same circumference?

Can you show a map that actually represents that? I'm having a hard time visualising...

How did you derive this formula?

It wasn't me. Through solving simultaneous equations.

Quote from: PizzaPlanet on February 16, 2011, 12:40:05 PM

Quote from: silver on February 16, 2011, 12:10:03 PM

The physical circumference of the south and north pole is the same. That is why it's relevant.

The geometry is not Euclidean. That's why it isn't.

Wait a minute.

[img ]http://i140.photobucket.com/albums/r36/Persistenxe/Flat_earth-1.png[/img]

The rim and the outer north pole have the same circumference?

Can you show a map that actually represents that? I'm having a hard time visualising...

IOA, either read before posting or don't post at all; preferably the latter. I've already explained the issue with the scale, and I'm tired of you being the permanent noob.

IOA, either read before posting or don't post at all; preferably the latter. I've already explained the issue with the scale, and I'm tired of you being the permanent noob.

Well thanks for explaining absolutely nothing. How am I supposed to know what you "already explained" if I don't even know what you mean?

What issue with scale? That the map doesn't correctly represent a flat Earth? If that's so, what kind of map would? If you were to travel high enough in the sky to see the "flat" Earth in its entirety, would you see a messed up Earth as well?

Well thanks for explaining absolutely nothing. How am I supposed to know what you "already explained" if I don't even know what you mean?

What issue with scale? That the map doesn't correctly represent a flat Earth? If that's so, what kind of map would? If you were to travel high enough in the sky to see the "flat" Earth in its entirety, would you see a messed up Earth as well?

You would know if you read the thread. These exact issues were already addressed here.

Using non-euclidean space and looking at the earth in a fractal manner we solve all issues with geography on a flat earth.

An excellent diagram, indeed!

I originally put it up on here in '06 and rarely post about it. Thats likely why you haven't seen it before. 

The rim and the outer north pole have the same circumference?

Can you show a map that actually represents that? I'm having a hard time visualising...

you know it works perfectly on a globe.....

Berny
Sailor, Astronomer, Chef

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

in this thread, I presented a logical proof of why light has nothing to do with physical distances, measured, real, or approximate.

"Logical" is a vast overstatement, as is "proof". However, light does have nothing to do with physical distances. I keep saying that all the time.
Mechanics != optics. Mechanics != optics. Mechanics != optics.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

And since you're not American

Ah, again with the nationality wildcard. Are you intentionally trying to make your arguments less credible?

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

I see how you could be confused about Colorado, but alas it was designed to be a perfect angular sector.

Alas, that was done under RE assumptions. It is completely inapplicable to the FE model.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

And the way you did your math is incorrect. That is not how you find surface area on a circle/disk.

Actually, it is very correct.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

The margin of error turns out to be like>5% if done correctly

Did you check my margin of error? It's slightly more than 2%.
Since I don't trust your mathematical capabilities (as they keep giving you completely wrong results over and over again), I'll do the calculations for you.
For reference:
The area of Colorado as given on Wikipedia: 269837 km^2
My half-assed approximation: 276012 km^2
[(276012-269837)/269837]*100%=~2.288%
[(276012-269837)/276012]*100%=~2.237%
2.288<5 and 2.237<5

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

If you'd allow me to show you

Humour me.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

The upper bound is 41 N and the lower is 37 N, the rate is 12,450 miles /180 degrees or 69.166666 miles per degree. Therefore, it is between 3389.16 miles and 3665.83 miles radially. Since its Long. coordinates are 102.03 and 109.03, the length is 7 degrees, or 0.122173048 radians.

And this is where you fail. You look like a map, which is an optical representation (and, as I stated before, gets more distorted as the scale diminishes), and treat it as if it could be used to make approximations. Sigh.
Mechanics != optics.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

thus the integral for the angular sector is

something we don't care about.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

Oh, come on. Surely you didn't think hotlinking a Wolfram output would work, did you?

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

or in other words, 119234 square miles, which is larger, as expected.

Or, in other words, you did some random inapplicable stuff and got a completely wrong result (14.44% margin of error!). How surprising. What's next? Proving that the Earth is round using the transfer function for a non-inverting operational amplifier?

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

Thusly, either
A, space itself is twisted in such a way that the earth is curved, yet "flat" aka bent space
B, the data from wikipedia is a part of the Conspiracy to trick you into thinking that the earth is not flat.
C, The Earth is not flat.

D, a^2+b^2=c^2 ergo camels are blue and the Earth is a spoon frequently used by Zeus. QED

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

A has been proven wrong, space has been found to be relatively flat.
B is still possible within the scope of this site.
C applying Occam's razor and common sense is most likely.

D is quite likely considering your use of inapplicable mathematics and your inability to understand one thing; and yes, I will say it again now; and yes, I will increase the font size again. Why? Because I want you to at least try to think about it.
Mechanics != optics.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

Note this does not prove the earth round

Definitely, seeing how it doesn't prove anything whatsoever.

Quote from: Thevoiceofreason on February 14, 2011, 07:27:46 AM

simply not flat and of the dimensions of the FAQ.

Definitely not, seeing how it doesn't prove anything whatsoever.

So in other words you have no idea what you're talking about. You seem to not understand how rectangles on polar coordinates work. Or how geometry works.

long. and lat. lines are the same in both. Why are longitude lines the same? It keeps the time zones the same. So yes, they are correct. You don't just multiply stuff together. you have to take the double integral over the area with respect to r dr dtheta. But you already knew that.

I assure you that I don't look like a map. And I used solely the coordinates in my approximation, with the knowledge that Colorado is a near perfect angular sector.

Is there any way to mathematically describe the surface of the FET model? For RET, we can very roughly approximate the surface area by using (4*pi*rē) where r is 6371km, right? So what is the FET equivalent?

He's just trying to troll at this point. Surface area on the flat earth has always maintained that long. and lat. coordinates where applicable. The surface area of the flat earth is pi*r^2

Using non-euclidean space and looking at the earth in a fractal manner we solve all issues with geography on a flat earth.

essentially you're saying that the earth has bendy space, that it exists in spherical space?
that is functionally equivalent to Round earth

Thought experiment. according to the bottom and top layers, walking along the south rim west, To your right side, is the known world. But according to the middle one, walking along the south rim going west, the known world is to your left

Quote from: Around And About on February 16, 2011, 01:20:49 AM

Is there any way to mathematically describe the surface of the FET model? For RET, we can very roughly approximate the surface area by using (4*pi*rē) where r is 6371km, right? So what is the FET equivalent?

Well it can't be pi * r^2, because that's for describing a flat, circular object, which according to PizzaPlanet it's apparently not anyways.

Let's state a fact and see what comes out of it.

Two circles with a different radius (IE distance from the North Pole) cannot have the same circumference.

In reality they do. The only geomitrically viable explanation for this is a sphere.

Or a circle in spherical coordinates that has teleporting loops on each end. Which happens to be perfectly equivalent to a round earth

Regardless of whether trolling is involved, it's fascinating to see the explanations involved, is all.

Regardless of whether trolling is involved, it's fascinating to see the explanations involved, is all.

Yes, and I too am interested in Johan Davis' diagram. This is inconsistent with the model I'm used to though. Why would the nudging birds argument happen if the earth was in bendy space?

Using non-euclidean space and looking at the earth in a fractal manner we solve all issues with geography on a flat earth.

What is this. I do not even.

Can you explain this diagram in more detail?

So in other words you have no idea what you're talking about. You seem to not understand how rectangles on polar coordinates work. Or how geometry works.

I assure you I do understand Euclidean geometries. They are simply inapplicable.

long. and lat. lines are the same in both.

Are they? It would seem you are attempting to inject definitions into my model. I'm sorry, but my model does not consider anything like latitude and longitude. They are redundant.

It keeps the time zones the same. So yes, they are correct. You don't just multiply stuff together. you have to take the double integral over the area with respect to r dr dtheta.

Again, you're adding stuff to my model. Stuff that directly contradicts the model. So, you're basically building a completely new and different model and then disproving it? I agree, your silly ideas don't work. You've managed to prove yourself wrong!

I assure you that I don't look like a map.

I am very certain you look like Colorado, considering your obsession.

And I used solely the coordinates in my approximation, with the knowledge that Colorado is a near perfect angular sector.

in RE*.

He's just trying to troll at this point.

Please stop stirring things up in this thread. The only person trying to troll here is you, with your "I WILL MAKE STUFF NO ONE SAID UP AND THEN PROVE THAT THIS STUFF IS WRONG, HEHEHEEHEHEHEHEHEHE" strategy.

Surface area on the flat earth has always maintained that long. and lat. coordinates where applicable.

This sentence doesn't make sense. I can't address its content, because it's inaccessible. You may want to revise your ("native", lol) English skills prior to posting.

The surface area of the flat earth is pi*r^2

Incorrect. Would you like to double integrate that and then apply Demorgan's Theorem? Oh, and once you're done with that, double integrate it again. You will find that this conclusively proves Colorado to be larger than Colorado. That's right, x>x. As you can see, RE is clearly impossible, because I just told you to do something that doesn't work, and it didn't work.

Thought experiment. according to the bottom and top layers, walking along the south rim west, To your right side, is the known world. But according to the middle one, walking along the south rim going west, the known world is to your left

It's both, you nitwit. Have you ever been to the Earth?

This is inconsistent with the model I'm used to though.

ITT: World in awe; Thevoiceofreason discovers that there are more FE models than 1.

all this talk is making it harder to understand your argument.  can you explain the main points so far?

what i got from the op:
light makes the outside circumfrence seem the same as the north pole
and google earth uses a flat projection of a sphere

light makes the outside circumfrence seem the same as the north pole

No, light makes it seem different whilst it is the same, and the optical illusion only occurs when we alter the scale of the map without altering its shape.

and google earth uses a flat projection of a sphere

And gets imprecise distances, for a similar reason FE does, yes.

That is impossible. When I say the known world, I mean the continents. They cannot possibly be on the left hand side of you, and on the right hand side.

That is impossible. When I say the known world, I mean the continents. They cannot possibly be on the left hand side of you, and on the right hand side.

They sure can, regardless of what direction you're facing...on a sssspherrrre. To be honest, I don't think that diagram could ever make any practical sense without some "doublethink" involved, although I'm sure this will simply be interpreted by some as "intellectual inflexibility" on my part.

Quote from: Thevoiceofreason on February 17, 2011, 03:40:35 AM

That is impossible. When I say the known world, I mean the continents. They cannot possibly be on the left hand side of you, and on the right hand side.

They sure can, regardless of what direction you're facing...on a sssspherrrre. To be honest, I don't think that diagram could ever make any practical sense without some "doublethink" involved, although I'm sure this will simply be interpreted by some as "intellectual inflexibility" on my part.

I'm having trouble understand your debunk. On the round earth, if you travel along the 89S line, to the west, you see the continents to your immediate right, as in the point to your right would be on 88.9999999... larger than graham's number of nine's..99998 S. you'd never see them on your left, unless you were at exactly 90 S. In which circumnavigation would become rotation, and the problem with this, is that you cannot nest another circle withing a single point.

I wasn't trying to debunk anything, just facetiously asserting that regardless of where you are or how you're oriented, you can either turn "right" or "left" (relative to your current direction) and eventually reach some land. On a globe, of course. Or EnglshGentleman's "teleporting edges" model.

I wasn't trying to debunk anything, just facetiously asserting that regardless of where you are or how you're oriented, you can either turn "right" or "left" (relative to your current direction) and eventually reach some land. On a globe, of course. Or EnglshGentleman's "teleporting edges" model.

oh I see I was trying to present a topological disproof based on my take of the diagram, which admittedly is confusing. It is rather trivial that one might turn left or right and walk far enough to reach land. I was talking about whether the world would be to your immediate left, in the mathematical sense. Some thing cannot be to your left and to your right in this sense, considering according to the bottom one, to your left is the north pole.

again, google maps is a flat map of a round earth. the distortion is proof of round earth.

again, google maps is a flat map of a round earth. the distortion is proof of round earth.

In your absence, they have invoked spherical geometry.

Quote from: wecl0me12 on February 17, 2011, 01:35:40 PM

again, google maps is a flat map of a round earth. the distortion is proof of round earth.

In your absence, they have invoked spherical geometry.

Spherical geometry on a disc, instead of spherical geometry on a sphere, ahh. Ockham ought to be rolling in his grave.

If you take a flat circle and call it Earth (what you state that the Earth is) with a radius of say 10 to keep it simple.

The center of this circle represents the north pole.

Now, let us measure the circumference of a new circle A within circle Earth that has a radius of 2. This new circle represents one trip around the north pole.

The trip around the northpole is pi*2^2 = 12,57 long.

Now, let us measure the circumference of another new circle B within circle Earth that has a radius of 8. This new circle represents one trip around the south pole.

The trip around the southpole is pi*8^2 = 201,06 long.

Point: 12,57 != 201,06

In the real world, the circumferences of these two circles A and B within circle Earth have been measured to be the same.

In a flat earth model these CANNOT be the same. If you say that this does not work like euclidean geometry, do please tell me why it does not, as that means you are assuming some weird kind of space warping and twisting to get what could be called "impossible" geometry, and certainly not a flat earth model.

Tl;dr, A and B cannot be the same on a flat earth yet they are, and euclidean principles that do not apply = impossible geometry = cannot be real. (See: Necker cubes and penrose stairs)

Quote from: vhu9644 on February 16, 2011, 11:23:06 PM

light makes the outside circumfrence seem the same as the north pole

No, light makes it seem different whilst it is the same, and the optical illusion only occurs when we alter the scale of the map without altering its shape.

Quote from: vhu9644 on February 16, 2011, 11:23:06 PM

and google earth uses a flat projection of a sphere

And gets imprecise distances, for a similar reason FE does, yes.

but if you took a perfect map of something, and you altered the scale, whithout changin the map, it would work fine right?

and google earth gets imprecise distances becuase it uses a flat projection of a sphere, not becuase the light is bending as it looks at it

Quote from: Thevoiceofreason on February 16, 2011, 07:22:28 PM

The surface area of the flat earth is pi*r^2

Incorrect. Would you like to double integrate that and then apply Demorgan's Theorem? Oh, and once you're done with that, double integrate it again. You will find that this conclusively proves Colorado to be larger than Colorado. That's right, x>x. As you can see, RE is clearly impossible, because I just told you to do something that doesn't work, and it didn't work.

i dont get what you are saying, his above quote is the surface area of a flat earth, and then you use the flat earth colorado, and it is bigger than colorado in re?

Quote from: PizzaPlanet on February 16, 2011, 11:27:20 PM

Quote from: vhu9644 on February 16, 2011, 11:23:06 PM

light makes the outside circumfrence seem the same as the north pole

No, light makes it seem different whilst it is the same, and the optical illusion only occurs when we alter the scale of the map without altering its shape.

Quote from: vhu9644 on February 16, 2011, 11:23:06 PM

and google earth uses a flat projection of a sphere

And gets imprecise distances, for a similar reason FE does, yes.

but if you took a perfect map of something, and you altered the scale, whithout changin the map, it would work fine right?

That's what I was also saying.  A map is an image.  If you resize it while maintaining the aspect ratio, you can take an image five square miles in size, resize it to one square foot, and the shapes of everything will look the same, only smaller with less detail.

Altering the scale of an image or map without altering it's shape will produce an image with the same shape, but different size.

Altering the scale of an image or map while altering it's shape will produce an image with a different shape and size than the original.