Distances on RE and FE consistent thanks to bendy light.

We are not concerned with the abstract and unobservable "curvature". The Earth is flat within the boundaries of reality. If you'd like to observe the Earth form outside the space-time continuum, join an arts club of some sort.

We are not concerned with the abstract and unobservable "intrisic curvature". The Earth is flat within the boundaries of reality. If you'd like to observe the Earth form outside the space-time continuum, join an arts club of some sort.

Intrinsic curvature would be observable. If you take a triangle on the defined surface, the angles will not add up to 180 degrees. That's observable. If you draw a large enough circle, its radius will not be equal to 2πR. That's observable.

Only if you take lines that are straight outside of the defined space, which will never occur in reality.

Only if you take lines that are straight outside of the defined space, which will never occur in reality.

I'll say it in a different way. If you take two crossing lines (geodesics) in non-Euclidean 3-space (let's make the lines perpendicular for ease of imagination), they will define a surface.

Everything I've said in the last to posts is happening in the non-Euclidean 3-space. I do not care how or even if it is embedded in some higher dimensional euclidean space.

Again, please learn more about manifolds.

Momentia, what's your view to the impossibility, according to top-notch scientist PizzazPlanet, to have a FE map with accurate distances and at any scale different to 1:1?

Everything I've said in the last to posts is happening in the non-Euclidean 3-space.

But it is viewed from outside of it, which is an impossibility.

I do not care how or even if it is embedded in some higher dimensional euclidean space.

Nor do I. In fact, you're the only person who keeps trying to "embed" [sic] something in a "higher dimensional" [sic] Euclidean space.

Again, please learn more about manifolds.

Ah, yes, the good old "You're clearly undereducated!" argument. It makes you ever so much more credible, doesn't it?

Momentia, what's your view to the impossibility, according to top-notch scientist PizzazPlanet, to have a FE map with accurate distances and at any scale different to 1:1?

I hold no such view.

Quote from: EmperorZhark on November 04, 2011, 05:41:53 PM

Momentia, what's your view to the impossibility, according to top-notch scientist PizzazPlanet, to have a FE map with accurate distances and at any scale different to 1:1?

I hold no such view.

Yes you do, you've been constantly telling that you couldn't produce an accurate 1:1 FE map and that because of some distorsion due to projection, your maps were accurate (or any kind of BS you have in your mind, for what I care) which clearly they aren't.

Yes you do, you've been constantly telling that you couldn't produce an accurate 1:1 FE map and that because of some distorsion due to projection

What? It is impossible to produce an accurate map of the Earth, be that round or flat, of any scale but 1:1.

your maps were accurate (or any kind of BS you have in your mind, for what I care) which clearly they aren't.

So you think a globe is not a good map of the Earth? Duly noted.

What? It is impossible to produce an accurate map of the Earth, be that round or flat, of any scale but 1:1.

I have a 1:1 map of the Earth in my closet.  I would show it to you, but it takes an eternity to open.  It is such a pain.  I wish the Earth was really round so I could just man-handle a globe instead.  Oh well.

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

Yes you do, you've been constantly telling that you couldn't produce an accurate 1:1 FE map and that because of some distorsion due to projection

What? It is impossible to produce an accurate map of the Earth, be that round or flat, of any scale but 1:1.

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

your maps were accurate (or any kind of BS you have in your mind, for what I care) which clearly they aren't.

So you think a globe is not a good map of the Earth? Duly noted.

1) Let's say 10% accurate
2) !?

Quote from: PizzaPlanet on November 05, 2011, 06:14:53 AM

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

Yes you do, you've been constantly telling that you couldn't produce an accurate 1:1 FE map and that because of some distorsion due to projection

What? It is impossible to produce an accurate map of the Earth, be that round or flat, of any scale but 1:1.

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

your maps were accurate (or any kind of BS you have in your mind, for what I care) which clearly they aren't.

So you think a globe is not a good map of the Earth? Duly noted.

1) Let's say 10% accurate
2) !?

1) I hope you mean 10% inaccurate...
2) That is not a proper response.

Quote from: EmperorZhark on November 05, 2011, 07:46:10 AM

Quote from: PizzaPlanet on November 05, 2011, 06:14:53 AM

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

Yes you do, you've been constantly telling that you couldn't produce an accurate 1:1 FE map and that because of some distorsion due to projection

What? It is impossible to produce an accurate map of the Earth, be that round or flat, of any scale but 1:1.

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

your maps were accurate (or any kind of BS you have in your mind, for what I care) which clearly they aren't.

So you think a globe is not a good map of the Earth? Duly noted.

1) Let's say 10% accurate
2) !?

1) I hope you mean 10% inaccurate...
2) That is not a proper response.

If you think a spherical map fits the Earth quite well with only slight inaccuracies, what do you those inaccuracies are?

Quote from: EmperorZhark on November 05, 2011, 07:46:10 AM

Quote from: PizzaPlanet on November 05, 2011, 06:14:53 AM

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

Yes you do, you've been constantly telling that you couldn't produce an accurate 1:1 FE map and that because of some distorsion due to projection

What? It is impossible to produce an accurate map of the Earth, be that round or flat, of any scale but 1:1.

Quote from: EmperorZhark on November 05, 2011, 05:39:38 AM

your maps were accurate (or any kind of BS you have in your mind, for what I care) which clearly they aren't.

So you think a globe is not a good map of the Earth? Duly noted.

1) Let's say 10% accurate
2) !?

1) I hope you mean 10% inaccurate...
2) That is not a proper response.

1) You're right
2) Yes, it is not, but PizzazPlanet is so far from understanding any basic stuff on maps that frankly I don't care.

Quote from: momentia on November 04, 2011, 04:24:24 PM

Again, please learn more about manifolds.

Ah, yes, the good old "You're clearly undereducated!" argument. It makes you ever so much more credible, doesn't it?

Pizza, you seem to be throwing around the term "non-euclidean" without understanding it.

To define a non-euclidean space of N dimensions, what other measurement do you need?

How do you construct your "flat plane" in said non-euclidean space?

Is there any way to tell the difference between a "flat plane" (in your notation) in euclidean vs. a "flat plane" in non-euclidean space if you are living on one of these planes?

Pizza, you seem to be throwing around the term "non-euclidean" without understanding it.

A non-Euclidean geometry is a geometry that isn't Euclidean. In other words, it doesn't follow all the axioms or "common notions" as described in the Elements.

To define a non-euclidean space of N dimensions, what other measurement do you need?

None at all.

How do you construct your "flat plane" in said non-euclidean space?

Your "two lines" idea works just fine.

Is there any way to tell the difference between a "flat plane" (in your notation) in euclidean vs. a "flat plane" in non-euclidean space if you are living on one of these planes?

Yes, assuming that by "euclidean space" you mean an Euclidean geometry, and by "non-euclidean space" you mean a non-Euclidean geometry. The flat plane wouldn't be flat (it wouldn't be possible at all, hence the "probably" - I'm making the assumption that the impossible request of yours is not impossible) in the abstract model known as "Euclidean geometry".

Quote from: momentia on November 05, 2011, 01:12:35 PM

Pizza, you seem to be throwing around the term "non-euclidean" without understanding it.

A non-Euclidean geometry is a geometry that isn't Euclidean. In other words, it doesn't follow all the axioms or "common notions" as described in the Elements.

Yes, in particular the fifth, the "parallel postulate," isn't followed.

Quote from: momentia on November 05, 2011, 01:12:35 PM

To define a non-euclidean space of N dimensions, what other measurement do you need?

None at all.

No, you need something called curvature. In 2 dimensional non-euclidean, this curvature is a constant number. If it is positive, all lines meet. If it is negative, for every line, and point not on that line, there is more than one line that is parallel to the first line passing through said point. When you define units of distance in such a geometry, you can give this curvature a number.

Quote from: momentia on November 05, 2011, 01:12:35 PM

How do you construct your "flat plane" in said non-euclidean space?

Your "two lines" idea works just fine.

Yes, and these to lines will defined a non-euclidean surface with 2 dimensions. It will thus have a curvature.

Quote from: momentia on November 05, 2011, 01:12:35 PM

Is there any way to tell the difference between a "flat plane" (in your notation) in euclidean vs. a "flat plane" in non-euclidean space if you are living on one of these planes?

Yes, assuming that by "euclidean space" you mean an Euclidean geometry, and by "non-euclidean space" you mean a non-Euclidean geometry. The flat plane wouldn't be flat (it wouldn't be possible at all, hence the "probably" - I'm making the assumption that the impossible request of yours is not impossible) in the abstract model known as "Euclidean geometry".

Yes, the two would be different. Your "flat plane" in non-euclidean geometry would still have the same metric as any other embedding of the surface. Assume constant h to get the metric of the "flat plane" at some height, Assume non-constant h to get the full metric of the earth (approximately):
h = height above sea level (in meters)
λ = longitude
Φ = latitude



By the way, this metric is based only on the surface in question, so it doesn't matter how you choose to visualize it. If distances on RE are correct, this metric is correct. But this metric also has curvature when you take h to be constant (take out the dh² term and set h to be constant), so the earth's surface isn't flat.

Yes, in particular the fifth, the "parallel postulate," isn't followed.

In this particular model, yes. In general, not necessarily.

No, you need something called curvature.

That is assuming that there is any.

Yes, and these to lines will defined a non-euclidean surface with 2 dimensions. It will thus have a curvature.

I don't see how it will. A flat surface has no curvature.

Yes, the two would be different. Your "flat plane" in non-euclidean geometry would still have the same metric as any other embedding of the surface. Assume constant h to get the metric of the "flat plane" at some height, Assume non-constant h to get the full metric of the earth (approximately):
h = height above sea level (in meters)
λ = longitude
Φ = latitude



By the way, this metric is based only on the surface in question, so it doesn't matter how you choose to visualize it. If distances on RE are correct, this metric is correct. But this metric also has curvature when you take h to be constant (take out the dh² term and set h to be constant), so the earth's surface isn't flat.

You assume that the Earth is round. It's not. I grow tired of you applying circular proofs to your claims. They all base on either the assumption that the geometry is Euclidean or that the Earth is an Euclidean surface within this non-Euclidean geometry.

Quote from: momentia on November 05, 2011, 04:24:28 PM

Yes, in particular the fifth, the "parallel postulate," isn't followed.

In this particular model, yes. In general, not necessarily.

Quote from: momentia on November 05, 2011, 04:24:28 PM

No, you need something called curvature.

That is assuming that there is any.

All non-euclidean geometries have curvature.

Quote from: momentia on November 05, 2011, 04:24:28 PM

Yes, and these to lines will defined a non-euclidean surface with 2 dimensions. It will thus have a curvature.

I don't see how it will. A flat surface has no curvature.

You assume that such a surface is flat because it is in euclidean geometry. In non-euclidean geometry, this is not the case.

Quote from: momentia on November 05, 2011, 04:24:28 PM

Yes, the two would be different. Your "flat plane" in non-euclidean geometry would still have the same metric as any other embedding of the surface. Assume constant h to get the metric of the "flat plane" at some height, Assume non-constant h to get the full metric of the earth (approximately):
h = height above sea level (in meters)
λ = longitude
Φ = latitude



By the way, this metric is based only on the surface in question, so it doesn't matter how you choose to visualize it. If distances on RE are correct, this metric is correct. But this metric also has curvature when you take h to be constant (take out the dh² term and set h to be constant), so the earth's surface isn't flat.

You assume that the Earth is round. It's not. I grow tired of you applying circular proofs to your claims. They all base on either the assumption that the geometry is Euclidean or that the Earth is an Euclidean surface within this non-Euclidean geometry.

**When I refer to the "earth's metric" in the following, I am referring to the metric shown with height 0. (h is 0, dh2 is gone).**

Incorrect, this metric is based on your claim that the earth's surface is isometric to a sphere (first order approx.) in euclidean space. That is your assumption, not mine. (Distances are the same on RE and FE)

Once a metric is defined (in any coordinate system) for a surface, the metric is the same for that surface embedded in any geometry. The metric determines distances distance along a path (on the surface) between two points. The metric is based on the surface and the coordinates chosen on that surface, not in any embedding.

The earth's metric has to be the same (using the same coordinates) for any surface that is isometric to the surface of the earth. (Note that this would be the same metric determined if I had decided to calculate the metric for the previously defined surface (two overlapping lines) in non-euclidean space.) However, ALL metrics for a surface give the same path length between two points along a given curve. (Different metrics come from different coordinate systems on the surface, but they all function the same.)

However, given ANY metric on a surface, curvature of that surface at any point can be uniquely determined.

Thus the earth's metric uniquely determines its curvatures, which happens to be non-zero. So the earth isn't flat.

Thanks for clearing things up.

All non-euclidean geometries have curvature.

Relative to what?

You assume that such a surface is flat because it is in euclidean geometry. In non-euclidean geometry, this is not the case.

In which non-Euclidean geometry?

Thus the earth's metric uniquely determines its curvatures, which happens to be non-zero. So the earth isn't flat.

The metric you have presented is of the Round Earth Theory. As I said: you assume that the Earth is round.
Keeping this in mind, I agree with you: If the Earth is round, it's not flat. In fact, I've admitted this several times in the past, and even in this thread.

Momentia's main issue is that he assumes he is the only person who knows what he's talking about, i.e. has ever taken a geometry class.

I was reading the 1st post on the 1st page of this topic and something occured to me:

Topography of a FE is a question of Euclidian geometry.
Where on flat geometry did the question of light (and time) ever occured?

These matters are not related. I understand that you use the notion of bendy light for sunsets, but for questions of distance, I don't see the relevance(as thousands of book on geometry do).

Quote from: momentia on November 05, 2011, 06:45:44 PM

All non-euclidean geometries have curvature.

Relative to what?

Intrinsically. "Curvature" means "Gauss curvature" here. Are you familiar with the Theorema Egregium ?

I was reading

Finally!

Where on flat geometry did the question of light (and time) ever occured?

You still have a lot of reading to do. Light has a whole lot to do with how the Earth looks and what we'd see on a map. It does have nothing to do with the physical shape of the Earth.

Intrinsically. "Curvature" means "Gauss curvature" here. Are you familiar with the Theorema Egregium ?

Ladies and gentleman, another valiant noob who doesn't read threads before posting!
Yes, Theorema Eregium has been brought up before, and, as I have explained, intrinsic curvature can only be observed from outside the geometry in question the only curvature that could be observed would be observable from outside the geometry in question. I cannot vouch for what the Earth looks like from outside the space-time continuum, nor do I care, since I believe there is no "outside of the space-time continuum".

Ok, you don't understand.

You'll never get me a FE map because you'll never admit that bendy light has nothing to do with maps, so you'll carry on with your useless theories about projection.

All your FE maps are wrong because they are RE maps.

And, yes, I've reading you post, what a waiste!

Ladies and gentleman, another valiant noob who doesn't read threads before posting!
Yes, Theorema Eregium has been brought up before, and, as I have explained, intrinsic curvature can only be observed from outside the geometry in question. I cannot vouch for what the Earth looks like from outside the space-time continuum, nor do I care, since I believe there is no "outside of the space-time continuum".

Another lack of understanding by pizza.

Intrinsic curvature is observable from inside the surface in question:

A curvature such as Gaussian curvature which is detectable to the "inhabitants" of a surface and not just outside observers. An extrinsic curvature, on the other hand, is not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides.

Metrics uniquely determine intrinsic curvature. not extrinsic curvature.

...and, as I have explained, intrinsic curvature can only be observed from outside the geometry in question.

Pardon me, but that's utter nonsense. Apparently you didn't understand the Theorema Egregium at all, as it states that curvature can actually be observed from inside the geometry in question. Hint: That's why it's called intrinsic.

By calling le a noob and then claiming that "intrinsic" means actually "not intrinsic", you are only making a fool of yourself.

And, yes, I've reading you post, what a waiste!

You've reading my post, huh?
You've learning English, or not waiste your time with it?

I've noticed that you don't read posts before responding to them, and that you tend to respond using horrible English. Perhaps the problem is that you simply don't understand what you're reading?

By calling le a noob and then claiming that "intrinsic" means actually "not intrinsic", you are only making a fool of yourself.

You are a forum noob, yes. Otherwise you'd know to read the thread before posting in it. As for me misspeaking, I am so incredibly sorry if I offended you. I have corrected my post, and am requoting my original response below.

To re-address the Theorema Egregium argument:

Quote from: PizzaPlanet on November 03, 2011, 12:21:47 PM

About rotundity: I have provided a distance-consistent model that is not spherical. You have made an assumption that distance-consistency implies rotundity. Confirmation bias proven.

"Mathematically speaking, a sphere and a plane are not isometric, even locally."
http://en.wikipedia.org/wiki/Theorema_Egregium

I know it's from wikipedia, but it is a valid result that follows from theorema egregium.

Two spaces are isometric iff there is an isometry between the two spaces.

An isometry is
"A bijective map between two metric spaces that preserves distances"
http://mathworld.wolfram.com/Isometry.html

This means if distances are consistent, the earth is not flat.

Within the same geometry, yes.
This is not the case.

Quote from: your link

in the ambient 3-dimensional Euclidean space.

Within the same geometry, yes.
This is not the case.

Quote from: your link

in the ambient 3-dimensional Euclidean space.

Quote from: momentia on November 03, 2011, 04:56:31 PM

Could you expand?

Why are you asking me to expand the same thing over and over again?
The principle you've posted applies to a three-dimensional Euclidean geometry.
I have already explained that the geometry of this model is not Euclidean.

We are speaking about Gauss curvature here, which is, according to the Theorema Egregium, intrinsic. Yes, the Theorema Egregium has already been mentioned, but apparently you still don't get it, as you keep babbling about embedding space and asking "relative to what" the curvature is defined.

Like it or not, Gauss curvature is intrinsic and thus independant from embedding space.

• If it's not zero, then the earth is not flat, as flat surfaces have Gauss curvature zero.
• If it's zero, then the earth is locally euclidean, hence it has an euclidean map. Please provide such a map.