You mean it fits the definition you provided to limit it to being flat.
As I've said, it is the only accurate determination of the geometry of Earth's surface, and it's straight and therefore flat.
No. My point is that it only works in Euclidean space (aka flat space). It does not work in non-Euclidean or non-flat space.
Euclidean space (or spaces which have Euclidean parts) are the only ones in which 2, straight, parallel lines remain the same distance apart.
I see what you mean here, but it doesn't rebut my point here, as I will explain.
It's pretty simple to see the relevance.
There is some line of "unknown" shape. (In reality we know the shape and we know it isn't straight.)
There is a reference line we can use, a "straight" line, or a geodesic in this curved space.
So, we move along the "straight" line, measuring the distance to the "unknown" line.
We find the "unknown" line remains the same distance.
According to your reasoning this means the "unknown" line, as it remains the same distance from a straight line, should be parallel to this straight line and thus straight.
The issue is that that only works in flat space.
Yes, I understand, and here's why it's not relevant.
What I'm claiming here is that if you have a line cross a surface, and you draw a straight line to the surface across the traversal at an equal distance, it will reveal the geometry of the surface. The key word here is "surface", I used that in the definition.
Your example with the wall 1000 km or so north of the equator you are traversing is able to get a parallel straight line geodesic parallel to a curved line, this is because in Non-Euclidean geometry (will represent a sphere here for demonstration), the line traversing the surface is parallel to a line straight across, which also change position in a third dimension relative to the parallel lines, which means this can only be represented in a 3D coordinate system.

This is required to define the curved and straight as parallel in this specific case.
Now, in defining the
surface, since it uses a line traversal, it only needs to be plotted on a 2D plane, and we can determine that in this case, the curved and straight lines being parallel doesn't hold.

This is because this isn't relying on this principle of non-euclidean geometry to determine a flat surface, so it doesn't hold and is irrelevant.
The 2 are equivalent.
The satellite, the geodesic through space time is equivalent to the equator.
The surface of Earth is equivalent to the line 1000 km north of the equator.
No it's not, since one doesn't rely on 3D non-euclidean geometry to determine this, the only reason why the curved and straight line can be parallel is because it is all plotted in an exclusively 3D coordinate system, it can't be represented in a 2D plane, like the surface geodesics can (these lines passing straight through the ground can be represented as 2D and straight parallel lines being the same distance apart holding.
In order for your method to work, the line 1000 km north of the equator must be a straight line.
It doesn't, there is a clear distinction.
BULLSHIT!
Any line can be used to define it.
Lets take a simple case of a sine wave, such as one of the form y=sin(x)+10.
We can use the x axis (y=0) as a reference.
In this case, the curve alternates between 9 and 11 units above it.
We can define it quite precisely.
So no, we do not need a line to remain the same distance.
That is your pathetic excuse to ignore all the other lines which show that Earth is not flat.
You need to actually justify your claim, not just spout more crap.
No, not any line. A Sine wave can't be congruent to a surface, If you use 9 and 11 unit alterations above the x axis, it could never match a line at y=1, so can't define the surface.
This method works because the lines are parallel in a plane, so they match the surface, and reveal it as straight.
No it isn't. It is the only way you can try to justify pretending Earth is flat.
And because of that, you will reject all the others because you think if they don't show Earth is flat they must be wrong.
It is literally the only way a line matches a surface, and it is straight and therefore a flat surface.
That doesn't address the point at all. I have also shown that that is not the case at all.
You failed to, you went on to an irrelevant 'gotcha' with non-euclidean geometry that misses how the surface is being determined here, I hardly call that 'showing it's not the case'.
For flat-space it is. Not for curved space.
Likewise, if you have a straight line which oscillates between 2 distances, that shows it cannot match the straight line and cannot be flat (which again is only true for flat space), and likewise, that is common sense. If the distance from a straight reference line is not a simple linear relation (e.g. y=mx+b), the line can't be straight.
But as I have said, it can only match the surface geometry it is parallel, an equal distance to match the changes in the surface directly, doing this with Earth (in non-euclidean space-time) reveals it's flat.
Again, they are parallel. There is a line going through all of them, which is perpendicular to all of them. That means they are parallel. So why don't they work?
Because the geodesic lines aren't parallel, ones traversing Earth's surface as a whole and not just straight down through.
So parallel as in not actually parallel and instead remaining the same distance apart. While the 2 are equivalent in flat space, they are not in non-flat spaces and in non-flat spaces, a line which remains the same distance away from a straight line will likely be curved.
I mean a line straight above a surface that can be represented in a 2D plane that are also able to be defined as a line whose points are at a fixed normal distance from a given surface, what you proclaim here doesn't apply to parallel surface geodesics.
Nothing in here indicates the motion is absolute rather than relative.
In order for this to be absolute the centre of the solar system needs to be fixed, absolutely.
No, it can be absolute in that the planets are orbiting the sun in an absolute sense.
Again, by accepting that the centre of the solar system can move, IT MEANS IT ISN'T ABSOLUTE!!! It isn't a difficult concept to grasp.
This is basically what you are saying:
"I accept that it doesn't need absolute motion, but relative to the reference frame of the solar system, it must be absolute (i.e. relative to the solar system)."
Notice the stupidity there?
It would be most simple to get what I mean by this statement:
The bodies in our defined solar system are orbiting the sun in an absolute sense.
Define objective.
It seems to be objective, not based upon any sentient entity, and it is capable of explaining the motions.
Happening independent of distinct reference frames that are being used completely to determine motion.
This is a statement, not an explanation.
You need to provide an explanation for why the planets move.
I'm just describing how they move in this case, relative to each other by means of reference frames. It's not just a statement, it's an explanation of motion with celestial bodies.
They move relative to each other because of Newton's first law in curved space-time with distinct reference frames for motion.
You don't need a coordinate system to define relative motion since relative motion is independent of any coordinate system imposed on them.
Stop trying to pretend objective and absolute mean the same thing.
You need to be able to describe the interactions between various entities to explain how objects move. You cannot do that from an Earth centred frame.
They do mean the same thing, it's absolute in that it doesn't depend on being between what's determined between reference frames. I describe the interactions as relative motions in curved space-time. Why they are in their specific positions instead of other positions will depend mostly on the geodesics they are present in with their state of motion.
No. I am saying it only works in flat space, not non-flat space. Even in 3D flat space, parallel lines point in the same direction.
In flat space a line needs 2 things to define it, a vector and a point. From that point you follow that vector (or the reverse) and you have your line. That vector is effectively the direction the line is pointing.
If you have 2 lines, which do not overlap, and have the same vector, they are parallel.
Exactly, and a straight line has the same vector as the Earth's surface and can be represented in a 2D plane as actually equivalent lines and not the non-euclidean 3D exclusive example you brought up which isn't relevant.
You seem to like going off on tangents to cover up your complete failure.
How about you stop setting up such pathetic strawmen and instead focus on what I have actually said?
As I have been doing.
Yes they are.
There is a line joining them which is perpendicular to each. They do not intersect. They are parallel. They just don't have the property of parallel lines which exists in flat space, that they remain the same distance. As they are no longer in flat space the distance between these parallel lines can vary.
Referring to this:
But don't worry, it gets more fun.
Now consider 4 objects above Earth. All heading due north (with no change in elevation). All above the equator at 0 degrees east. All at the same velocity. But now one is in a perfectly circular orbit, while another is significantly below it, in an elliptical orbit, with the other 2 above, one in an elliptical orbit and one in a hyperbolic orbit.
All 4 of these lines are currently parallel because they are perpendicular to the line connecting them, but let them move on, and their paths are completely different.
Only, as I have shown, the paths with the surface can be represented in 2D plane, and when you do that, the lines are not parallel as they are not an equal distant apart, and when you translate the surface geodesics to a plane, they clearly aren't parallel, in fact, if you have an elliptical orbit that gets lowest to the Earth at two points that are say x distance above and a circular orbit at x distance above constantly, they will intersect, which clearly can't be parallel lines.
So, it doesn't work, only one parallel to Earth's surface, equidistant from it at all points, to match the surface. If you have an elliptical orbit, the surface and the geodesic get's closer, so under this method, an undulating surface could only be determined if a path that never got closer to Earth revealed an undulating path, but only a straight path does this in the case of Earth in curved space-time.
No. I meant flat, i.e. Euclidean, i.e. parallel lines remain the same distance apart; there is only a single straight line through a point, parallel to another straight line; the sum of the internal angles of a triangle is 180 degrees; etc.
This is different from non-flat spaces, where instead parallel lines can converge or diverge; there can be none (depending on the definition), 1 or infinitely many "straight" lines through a point, parallel to another straight line; the sum of the internal angles of a triangle can be more or less than 180 degrees.
Stop trying to pretend I am saying something else. Start addressing what I have actually said.
Only, the moment we transfer this to the plane representation of the line cutting to the surface (the line doesn't have width so it doesn't require 3D to represent), we reveal that the straight line reveals a flat surface. Parallel lines can't converge since they are an equal distance apart constantly by definition, their vectors are the same.
I'm not pretending either, I legitimately and unintentionally misunderstood.
That is because you don't want to admit you were wrong.
Believe it or not, I still fail to see how I'm wrong.
It shows how you can have a line remain equidistant from a straight line, yet not be straight.
This is because it is not in flat space.
Not in this case, you have to look at why this is the case in the non-euclidean example you used, and when you have that and apply it to the surface geodesics, it is revealed that they are not connected and therefore irrelevant.
No it doesn't.
Again, that requires the space to be flat. If you don't have flat space that entire line of reasoning is baseless. It relies upon a fact of mathematics which is only valid for flat spaces, that parallel lines always remain the same distance apart.
In my case, Earth is a flat surface since an equidistant line traverses it as straight, the non-euclidean nature of space-time isn't quite relevant in this case (other than that a straight line meets itself because it's in non-euclidean space) since the line has no width, therefore, it can be represented in 2D, and from there it is determined that parallel to this line is a flat surface.
Yes, it is the same, by which I mean completely false. But this does help show that.
The fact that you get parallel geodesics which intersect each other shows that 2 lines do not need to remain the same distance to be parallel and thus straight.
It shows that the distance between straight lines varies, and thus you can have a line which is not straight remain the same distance from a line which is straight.
Then show that this is the case with
my specific point here with a line parallel to a surface of earth (You will only need to represent it in 2D since the lines have no width, just distance from the surface and length across) and I will abandon this model.